A critical skill you must have is the ability to write and balance equations. So much of the heart of chemistry is based on the balanced equation that the study of chemistry is impossible without this skill. The ability to balance an equation is a great deal easier than the skills needed to balance the family checkbook, but the skills required are the same. You can’t create money, nor can you ignore it, when you are balancing a checkbook. The chemistry version is the law of conservation of matter: matter can neither be created nor destroyed by ordinary chemical means. Just as a bank is required to keep track of the movement of money with no gain or loss of money, the chemist must account for all matter originally present, the reagents, and where that matter goes, the products, with no gain or loss of matter. The chemist’s actions produce a balanced equation, coefficients specifying the number of molecules (or formula units) of each species involved. The coefficient is the number in front of the participants in a chemical reaction that tells us how much of the substance is present. The resulting balanced equation is an expression of the total mass of the reagents being equal to the total mass of the products.
The relative number of reacting and resulting molecules are indicated by the coefficients associated with the molecules. For example, the combustion of ammonia in oxygen is described by the balanced chemical equation
which can be read like a sentence as four molecules of ammonia react with three molecules of oxygen to yield/produce/form two molecules of nitrogen and six molecules of water. If we think of an equation being a recipe, then the coefficients tell us how much of the ingredients (the reagents or reactants) we need and the amount of the products that will be produced. Further, the arrow (yields sign) implies that the reaction may go to completion (all the way to the right) if the amount of the reactants conforms to the 4:3 ratio of molecules stated. Incidentally, the ratio provided in molecules in this discussion could just as easily have been presented in moles, could it not?
Some reactions between chemical substances occur almost instantaneously on mixing; some go to completion after sufficient time has elapsed; and some reactions go to only a partial extent even after a lot of time has gone by. Just about the only way to determine the nature of the reaction is to perform it empirically (in the laboratory). Even without running the reaction, we can interpret the balanced equation by stating that, if a large number of ammonia and oxygen molecules are mixed, a certain number of nitrogen and water molecules will be formed. At a given instant it is not necessary that the NH3 or O2 is all consumed, but whatever reaction does occur takes place in the molecular (or mole) ratio prescribed by the equation.
In the above balanced reaction, the atoms in seven indicated molecules on the left rearrange to form eight molecules on the right. There is no algebraic rule governing these numbers of molecules, but the number of atoms on each side of the equation must balance for each element, since the reaction obeys the law of conservation of matter, as mentioned above. The number of atoms of any element occurring in a given substance is found by multiplying the subscript of that element in the formula by the coefficient of the formula. The addition of specific atoms in the equation tells us that there are 4 nitrogen atoms both left and right (4NH3 → 2N2), 12 hydrogen atoms on both sides (4NH3 → 6H2O), and, likewise, 6 oxygen atoms (3O2 → 6H2O).
Because one mole of any substance is a specific number of molecules (1 mol things = 6.02 × 1023 things from Chapter 2), the relative numbers of moles undergoing reaction are the same as the relative numbers of molecules. Because of the relationship of molecules to moles, the equation above can be interpreted in terms of masses calculated directly from the Periodic Table (H = 1, O = 16, N = 24, all in g/mol).
The equation shows that 4 mol NH3 (4 mol × 17 g/mol) reacts with 3 mol O2 (3 mol × 32 g/mol) to form 2 mol N2 (2 mol × 14 g/mol) and 6 mol H2O (6 mol × 18 g/mol). More generally, the equation shows that the masses are in the ratio of 68:96:56:108 (or 17:24:14:27 with all common factors removed). The mass ratio is the same regardless of which mass unit is used (g, kg, lb, ton, etc.).
The equation with which we have been working is an example of what happens in all cases: The law of conservation of mass requires that the sum of the reactant masses (68 + 96 = 164 units) be equal to the sum of the resultant product masses (56 + 108 = 164 units).
The importance of mass relations when dealing with chemical equations may be summarized as follows:
1. Mass relations are as exacting as the law of conversation of mass (matter).
2. Mass relations do not require any knowledge about the variable conditions; for example, whether the H2Ois in the liquid form or is steam.
3. Mass relationships do not require any knowledge of the true molecular formulas. In the above example the masses or the number of atoms would be unchanged if the oxygen were assumed to be ozone (2O3, instead of 3O2). In either case, the equation would be balanced with 6 oxygen atoms on each side. Similarly, if the water molecules were polymerized, mass relations would be the same whether the equation contained 6H2O, 3H4O2, or 2H6O3. This principle is very important in cases where the true molecular formulas are not known. Mass relations are valid for the many equations involving molecules that may dissociate (S8, P4, H6F6, N2O4, I2, and many others) or those that associate to form complex polymers, such as the many industrially important derivatives of formaldehyde, starch, cellulose, nylon, synthetic rubbers, silicones, etc., regardless of whether empirical or molecular formulas are used.
When we are provided with the mass of one of the reactants, we normally assume that the other reactants are in sufficient supply to react or are in excess. What happens if we are given amounts of more than one reactant? We are then responsible for determining if there is a shortage of one or more reactants because the reaction stops when that reactant has been used up. The reactant in the least supply is referred to as the limiting reactant and calculations producing the expected amount of products are performed on the basis of the limiting reactant. All reactants in this situation that are not the limiting reactant are the excess reactants. See Problems 4.7 and 4.8 for examples of limiting reactant problems.
Skill in balancing equations will increase very quickly with practice, especially as you learn to recognize the various types of chemical reactions. Once you have recognized the types of chemical reactions, you will be able to predict the products if you are given just the reactants. A few examples of the more predictable types of reactions are listed below.
1. Combustion reactions. Oxygen in excess (usually from the air) combines with organic compounds, those composed of carbon, hydrogen, oxygen, and possibly other elements. Because of the presence of carbon and usually hydrogen, carbon dioxide and water are expected to be products, as the burning of nonane (C9H20) shows.
2. Replacement (displacement) reactions. A more active element can replace a less active one in a compound.
3. Double displacement (metathesis) reactions. This reaction commonly occurs in solution when the reactants produce ionic solution with an exchange of ions if one combination produces a compound that precipitates an insoluble salt.
4. Acid-metal reactions. An acid, such as HCl, HF, H2CO3, and a metal more active chemically than the acid’s hydrogen can react to form a salt and hydrogen gas.
5. Acid-base reactions (neutralization). An acid, which contributes H+ (H3O+) ions, and a base, which contributes OH– ions, undergo metathesis to produce water (HOH or H2O) and a salt. Isn’t this a special case of a double displacement reaction?
With reference to our use of HOH: Balancing the water in these reactions can be a problem since the hydrogen found in the water comes from different sources, hydrogen and hydroxide ions. Balancing the hydrogen ion (hydrondium ion) against the hydrogen in water and the hydroxide on the left against the hydroxide on the right simplifies the balancing process. Try balancing the reaction of nitric acid and magnesium hydroxide using H2O on the right and this problem will become clear.
6. Combination reactions. Elements and/or compounds combine into one product.
The reaction involving P4 and Cl2 depends on the ratio of the reactants, temperature, and pressure.
7. Decomposition reactions. A single reactant is transformed by heat or electricity into two or more products.
The reaction involving HgO depends on temperature and oxygen pressure.
4.1. Balance the following equations:
(a) Li + ZnCl2 → Zn + LiCl
(b) FeS2 + O2 → Fe2O3 + S2O
(C)C7H6O22 + O2 → CO2 + H2O
(a) There are no fixed rules for balancing equations. Often, a trial-and-error procedure is used (try a coefficient and replace if incorrect). The term trial-and-error doesn’t mean that there isn’t any order or pattern. If you choose a plan of action and follow it consistently, the balancing of equations becomes straightforward, if not easy. A most logical starting place is to look at the left side of the equation and choose an element to balance against the right side.
Let us start with the element furthest on the left, lithium. Since there is one atom on the left and also on the right, we can consider lithium balanced—this might change. The next element is zinc, with one atom on both the left and the right. Chlorine, though, is not in balance because there are two atoms on the left, but only one on the right. Doubling the chlorine on the right will balance the chlorine.
We now have an error which will be recognized when we go through the balancing procedure again. There is one lithium on the left, but there are two on the right. Doubling the lithium on the left will balance the lithium.
Checking further using the left-to-right pattern as above, we find that the zinc and the chlorine are in balance; therefore, the equation is balanced. It wouldn’t hurt to go through the check process one more time just to make certain.
One of the common misconceptions is that we are “putting a 2 in front of the LiCl” in the first step and “putting a 2 in front of the Li” in the second step. This is incorrect; we are multiplying by the coefficient in each case, not just putting it there. This may sound picky, but it is important since the balancing process can become confusing with the more complex chemical reactions. Understanding the process is important, not just going through the motions.
(b) This equation is a little more complex because the oxygen appears in one place on the left and is found in two places on the right. Even so, we can use the same pattern just discussed by starting with the left-most element.
There is one iron on the left, but there are two on the right. Let’s double the iron on the left.
The next element is sulfur with 4 on the left and only 1 on the right. If we multiply the SO2 by 4, we will have balanced the sulfur. Notice that the 4 sulfurs on the left are counted because there are two FeS2 molecules providing 2 × 2 = 4 sulfur atoms on the left of the equation.
The last element to be balanced is the oxygen. The oxygen on the left can be multiplied by a coefficient; however, the amount of the oxygen on the right has been established by balancing the iron and sulfur. There are eleven (4 × 2 + 3 = 11) oxygen atoms on the right. Multiplying the oxygen on the left by 5
will balance the oxygen.
A final check of the elements tells us that the equation is now balanced. There is a potential problem with this equation as balanced. Most often, equations are balanced without the use of fractions. We can eliminate the fraction () by multiplying the participants in the reaction by 2.
The equation is still balanced because the ratio of participants in the reaction has been preserved since we multiplied all participants by the same number, 2.
(c) Carbon is the left-most element in the equation.
Let us multiply the carbon in carbon dioxide by 7. Since we can’t change the ratio within the compound, we multiply the entire molecule, CO2, by 7.
The next element is hydrogen. Multiplying the water by 3 will provide us with the 6 hydrogen atoms we need to balance the equation to this point.
There are 2 oxygen atoms on the left in C7H6O2, but the amount of oxygen in O2 just to the left of the yields sign (arrow) is not fixed because we can multiply the molecule as needed. On the right side of the equation, we have set 17 oxygen atoms (7 × 2 + 3 = 17). If we subtract the 2 oxygens in C7H6O2 on the left from 17, we must account for 15 more oxygens on the left to balance against the right side of the equation. Multiplying O2 by 7 will do the job.
And, if we were to double the coefficients in the equation, we will have balanced the equation using whole numbers, as is traditional.
4.2. Complete and balance the following equations that occur in aqueous solution (water is the solvent). [Note: Barium phosphate, Ba3(PO4)2, is very insoluble; tin is a more active metal than silver.]
(a) Ba(NO3)2 + Na3PO4→
(b)Sn + AgNO3 →
(c)HC2H3O2 + Ba(OH)2 →
(a) Writing the products of the reaction requires the recognition of a metathesis (double displacement) reaction. The reason why this reaction progresses to the right is that the barium phosphate is insoluble and the sodium salts are soluble in water. The precipitation of barium phosphate essentially removes it from the reaction environment.
Once we have established the products and start checking atoms from the left of the equation, we notice that the barium is out of balance. Tripling the barium nitrate on the left will fix that issue.
Notice that the nitrogen in nitrate, 
, does not appear twice on either side of the equation. This means that the nitrate ion does not come apart; we can balance nitrogen by balancing nitrate. This balancing act requires that we multiply the nitrate in sodium nitrate on the right by six.
The next unit to balance is the sodium. There are 3 on the left and there are 6 on the right. If we were to double the sodium on the left (2 × 1Na3PO4), we will have balanced the sodium.
Lastly, we look at the phosphate ion, 
and note that phosphate is already balanced.
We should always run through the equation one more time to make certain of the balancing and, when we do so for this reaction, we find that it is balanced.
(b) Since we have been informed that tin is a more active element than is silver, we can expect tin to replace the silver in silver nitrate in a single replacement reaction.
Working from the left, we find that the tin is currently in balance; then we note that the silver is also. However, the nitrate ion is definitely not balanced. Let’s double the silver nitrate on the left.
We double check our work and find that although the tin is OK, the silver is not. There are now 2 silver atoms on the left and only 1 on the right. If we double the silver on the right, it will be balanced.
Another check of the equation tells us that the tin is balanced with 1 on each side, the silver is balanced with 2 each side, and the nitrate is also balanced with 2 on each side. The equation is balanced.
We had to go through this equation 3 times to balance and check. A hint that there might be a problem in balancing using the trial-and-error system is having to change the numbers when you are going through a balancing/checking procedure for the third time. If this happens, look to make certain that all of the participants are written correctly as there are ways of incorrectly writing formulas that will make balancing the equation impossible. Unfortunately, some equations can be balanced with incorrectly written formulas.
(c) Recognizing that HC2H3O2 + Ba(OH)2 is an acid-base reaction, you can predict the products by recalling that acid-base reactions produce water and the remaining ions form a salt. The acid-base reactions proceed because water does not ionize, which drives the reactions to the right.
First, notice that water has been written as HOH, rather than H2O, because the hydrogen is coming from the acetic acid and also from the barium hydroxide as part of the hydroxide ion, OH–. Another factor to note is that acetic acid can be written as CH3COOH; however, we choose to write the acid in the form matching the other acids (inorganic acids) with the hydrogen that is giving the acid its character on the left (HCl, HNO3, H2SO4, etc.). The last observation to be made is that the hydrogen on the left in HC2H3O2 will be balanced with the left-hand hydrogen in HOH; similarly, the OH in Ba(OH)2 is the only source of hydrogen and oxygen balanced with the OH in HOH.
Starting with the hydrogen in the acid, we notice that it is in balance with the hydrogen in the water. However, the acetate ion, 
on the left is not in balance with the acetate on the right. Doubling the acetate on the left will balance the ion.
Continuing, the barium is in balance; the OH is not. Doubling the water balances the hydroxide.
A check of the balancing show us that the equation is in balance.
4.3. Caustic soda, NaOH, is often prepared commercially by the reaction of sodium carbonate, Na2CO3, with slaked lime, Ca(OH)2. How many grams of NaOH can be obtained by treating 1 kg of Na2CO3 with Ca(OH)2?
Most of the time that you can write an equation to describe what is happening, it is a good idea to do so. The equation is essentially a set of instructions telling us how to perform a process. Further, the equation tells us how much of each of the reactants to use and how much to expect of the products. These factors are provided by the coefficients and can be considered in moles of substances. If we have moles of substances, we can easily convert to grams of substances using the atomic masses of the atoms involved provided by the Periodic Table of the Elements. The equation for this reaction is
We are only interested in the Na2CO3 and NaOH and no longer need to concern ourselves with the Ca(OH)2 and CaCO3 due to the way in which the problem is worded. However, we could not have arrived at the mole ratio of 1:2, nor the mass ratio, 106.0/80.0, without the balanced equation. Depending on how we work the problem, one or both of the relationships is critical to arriving at the correct answer.
Mole Method: As in Chapter 2, the symbol n(X) will be used to refer to the number of moles of substance whose formula is X and m(X) will denote the mass of substance X. Let’s consider 1000 g Na2CO3.
From the coefficients in the balanced equation, n(NaOH) = 2m(Na2CO3) = 2(9.434) = 18.87 mol NaOH.
Proportion Method: This method applies the information established by the balanced equation directly. The best part of this method is that the information is presented in a logical manner that practically solves the problem for you. The trick is to write the information given by the balanced equation (Eq. Info.) above the equation and the information provided by the problem (Prob. Info.) below the equation. Write the information above and below the participant identified in the equation by the problem, ignoring those participants not required for the solution of the problem. Place a symbol representing the unknown value as needed in Prob. Info. (W reminds us that the answer should come out in weight—grams.)
We next must consider that the units are different and, regardless of what we are to do, the units should be the same—mol and kg must become either moles or kilograms. The units of grams are the desired answer so that we could convert from kilograms as requested by the problem (1 kg = 1000 g).
We are now ready to set up a ratio and proportion (2 fractions equal to each other) and solve for W.
Rearrangement solves for W and the weight of the NaOH produced.
Had we noticed that the g Na2CO3 would cancel out just above, we could have done so before the rearrangement and saved a little writing. Either way, it is important to note that the solution tells us that the units on the answer are to be in grams NaOH, just as was required by the problem.
4.4. The equation for the preparation of phosphorus in an electric furnace is
Determine
(a) the number of moles of phosphorus formed for each mole of Ca3(PO4)2 used;
(b) the number of grams of phosphorus formed for each mole of Ca3(PO4)2 used;
(c) the number of grams of phosphorus formed for each gram of Ca3(PO4)2 used;
(d) the number of pounds of phosphorus formed for each pound of Ca3(PO4)2 used;
(e) the number of tons of phosphorus formed for each ton of Ca3(PO4)2 used;
(f) the number of moles each of SiO2 and C required for each mole of Ca3(PO4)2 used.
(a) From the equation, 1 mol P4 is obtained for 2 mol Ca3(PO4)2 used, or, mol P4 per mole of Ca3(PO4)2.
(b) The molar mass of P4 is 124. Then mol P4 = 2 × 124 = 62 g P4.
(c) One mole Ca3(PO4)2 (319 g/mol) yields mol P4 (62 g). The 1 g Ca3(PO4)2 gives 62/310 = 0.20 g P4.
(d) 0.20 lb; the relative amounts are the same as (c) regardless of the units.
(e) 0.20 ton (as above).
(f) From the coefficients in the balanced equation, 1 mol Ca3(PO4)2 required 3 mol SiO2 and 5 mol C.
4.5. Hydrochloric acid was once prepared commercially by heating NaCl with concentrated H2SO4. How much sulfuric acid containing 90.0% H2SO4 by weight is needed for the production of 1000 kg of concentrated hydrochloric acid containing 42.0% HCl by weight?
(1) Amount of pure HCl in 1000 kg of 42.0% acid is 0.420 × 1000 kg = 420 kg.
(2) We need the balanced equation and the molecular masses to determine the answer.
The relationship from the balanced equation is 1 mol H2SO4 (1 × 98.1 = 98.1 g) is required to produce 2 mol HCl (2 × 36.46 = 72.92 g). Therefore,
(3) Finally, we determine the amount of sulfuric acid solution containing 90.0% H2SO4 that contains 565 kg of pure H2SO4. We now know that 0.900 kg of pure H2SO4 makes 1 kg of 90.0% solution. Then
Factor-Label Method: This method has the advantage of allowing you to write down the solution as one step. There is also the added advantage of having the calculation laid out ready to be put through your calculator. As you read this setup from left to right, cross out the units that cancel and notice that the units on the answer are those that are to be on the final answer.
4.6. Before the public became serious about pollution, it was common to improve the performance of gasoline by the addition of lead compounds. A 100-octane aviation gasoline used 1.00 cm3 of tetraethyl lead, (C2H5)4Pb, of density 1.66 g/cm3, per liter of product. How many grams of ethyl chloride, C2H5Cl, are needed to make enough tetraethyl lead for 1.00 L gasoline? Tetraethyl lead production is by
The mass of 1.00 cm3 (C2H5)4Pb is (1.00 cm3)(1.66 g/cm3) = 1.66g required per liter. In terms of moles tetraethyl lead needed per liter:
The chemical equation tells us that 1 mol (C2H5)4 Pb requires 4 mol C2H5Cl. Using this information, we find that 4(0.00514) = 0.0206 mol C2H5Cl is needed. And,
Factor-Label Method: Let us use the nonstandard abbreviation TEPb for tetraethyl lead as a convenience.
4.7. A solution containing 2.00 g Hg(NO3)2 was added to a solution containing 2.00 g Na2S. Calculate the mass of the insoluble HgS formed according to the reaction, Hg(NO3)2 + Na2S → HgS + 2NaNO3.
This problem gives us the amounts of both the reactants; this is a hint that we might be dealing with a limiting reactant problem. In other words, one of the reactants might be used up before the other and that will stop the progression of the reaction. Further, since the mass of both reactants is 2.00 g, the presence of a limiting reactant is nearly guaranteed.
We can use the proportion method to set up the problem and test to determine which of the reactants, if any, is in short supply. Let’s use t as the symbol to represent the test amount of a reagent. As has been mentioned, a consistent pattern of approach is a good way to avoid confusion; we test the second reactant to hold to pattern. Instead of using the 2.00 g of Na2S given in the problem, let t represent the amount of Na2S needed to just react the 2.00 g Hg(NO3)2 and see if we were given enough. Additionally, let’s do the calculation in grams as given in the problem. In order to do so, we must express the Eq. Info. in terms of grams.
Once we have the values, we set up the ratio and proportion, then solve for t. This time, we decided to cancel out like units as we brought them down from the equation to the ratio and proportion (shortcut).
The interpretation is pretty straight forward because we just found out that we need 0.48 g Na2S (the value of t)to use up the 2.00 g Hg(NO3)2 given in the problem. We have a good deal more of the Na2S than we need, approximately one and a half grams too much to use up the other reactant. The limiting reactant is the mercury(II) nitrate, Hg(NO3)2. We must go back to the relationship of the mercury(II) nitrate to the desired product, mercury(II) sulfide. Once the equation has been balanced, we can ignore the NaNO3 since the problem requested no information about it. A is being used as the symbol to represent the amount of HgS that will be produced.
We use the information associated with the balanced equation to set up the ratio and proportion. Again, we need to watch to see if the desired unit appears in the solution.
In summary, the original 4.00 g of reagents has been transformed to 1.43 g HgS product and there is an excess of 1.52 g Na2S (2.00 g – 0.48 g from the test).
4.8. How many grams of Ca3(PO4)2 can be made by mixing a solution containing 5.00 g calcium chloride with another containing 8.00 g potassium phosphate? The reaction is
Using the same procedure as in Problem 4.7, the first action is to run the test to see if there is a limiting reactant. We set up the balanced equation with the information we were given and test to see if there is sufficient K3PO4 to use up all of the CaCl2.
Canceling similar units as the values are brought from the equation to the ratio and proportion gives us
Since 6.37 g K3PO4 are required to use up all of the calcium chloride and we were given 8.00 g K3PO4, the limiting reactant is the CaCl2. We are now able to set up the balanced equation with the equation information and the problem information. Since we know that the limiting reactant is the CaCl2, that we have been directed to calculate the amount of Ca3(PO4)2, and that the ratio of CaCl2 to Ca3(PO4)2 is 3:1, we can set up the ratio and proportion and solve for the answer to this problem.
Notes: (1) If there had been no limiting reactant, it would not matter which of the reactants were used for the final calculations. (2) It does not matter which of the reactants you chose for the test (t) so long as you are able to correctly interpret the results.
4.9. In one process for waterproofing, a fabric is exposed to (CH3)2SiCl2 vapor. The vapor reacts with hydroxyl groups on the surface of the fabric or with traces of water to form the waterproofing film [(CH3)2SiO]n by the reaction
where n stands for a large integer. The waterproofing film is deposited on the fabric building up layer on layer. Each layer is 6 Å thick, the thickness of the (CH3)2SiO group. How much (CH3)2SiCl2 is needed to waterproof one side of a piece of fabric that is 1 m by 2 m with a film of 300 layers thick? The density of the film is 1.0 g/m3.
Note: The unknown integer n canceled in the factor-label calculation along with all units except grams.
4.10. What is the percent free SO3 in an oleum (considered as a solution of SO3 in H2SO4) that is labeled “109% H2SO4”? Such a designation refers to the total weight of pure H2SO4, 109 g, that would be present after adding sufficient water to 100 g of the oleum to convert it to pure H2SO4.
Nine g H2O will combine with all the free SO3 in 100 g oleum to give a total of 109 g H2SO4. The equation H2O + SO3 → H2SO4 indicates that 1 mol of H2O (18 g) combines with 1 mol SO3 (80 g). Then
Therefore, 100 g of the oleum contains 40 g SO3, or the percent free SO3 in the oleum is 40%
4.11. KClO4 may be made by a series of sequential reactions. How much Cl2 is needed to prepare 100 g KClO4 by the sequence?
The mole method and the factor-label method are the simplest routes to the solution of this problem. In neither case is it necessary to find the masses of the intermediate products.
Mole Method
Factor-Label Method
4.12. A 1.2048-g sample of Na2CO3 is suspected of being impure. It is dissolved and allowed to react with CaCl2. The resulting CaCO3, after precipitation, filtration, and drying, was found to weigh 1.0262 grams. Assuming that the impurities do not contribute to the weight of the precipitate, calculate the percentage purity of the Na2CO3.
One way of working this problem is to calculate the amount of sodium carbonate that was used in the reaction, then compare that weight to the weight of the sample. Since the problem tells us that there has been a reaction, an equation is written to start us on the way to solving the problem. We may also write the information provided by the problem and by the equation itself to find out just how much Na2CO3 was in the sample (S weight of Na2CO3). This setup also provides us with the weights we need to calculate the percent composition Na2CO3.
Since the sample weight was 1.2048 g and the actual amount of Na2CO3 present is 1.0867 g, the calculation is as 1.0867 g Na2CO3 in sample
4.13. A mixture of NaCl and KCl weighed 5.4892 g. The sample was dissolved in water and silver nitrate was added to the solution. A white precipitate was formed, AgCl. The weight of the dry AgCl was 12.7052 g. What was the percentage NaCl in the mixture?
Silver ions react with chloride ions and those chloride ions are supplied by both of the salts.
By the law of conservation of matter, we must account for all of the chlorine atoms; there are two sources of chlorine. Therefore, the total amount of chlorine formed is the sum of that formed in the two equations. Since the chlorine in both reactions is tied up as AgCl and there is one mole of chlorine per mole of AgCl,
Let y = mass of NaCl and z = mass of KCl. Then
A second equation for the unknown masses is provided by the data:
Eliminating z between (1) and (2), and solving for y, we obtain y = m(NaCl) = 4.0624 g. Then,
Balance the following equations:
4.14. C2H4(OH)2 + O2 → CO2 + H2O
Ans. 2C2H4(OH)2 + 5O2 → 4CO2 + 6H2O
4.15. Li + H2O → LiOH + H2
Ans. 2Li + 2H2O → 2LiOH + H2
4.16. Sn + SnCl4 → SnCl2
Ans. Sn + SnCl4 → 2SnCl2
4.17. Ba(OH)2 + AlCl3 → Al(OH)3 + BaCl2
Ans. 3Ba(OH)2 + 2AlCl3 → 2Al(OH)3 + 3BaCl2
4.18. KHC8H4O4 + KOH → K2C8H4O4 + H2O
Ans. KHC8H4O4 + KOH → K2C8H4O4 + H2O
4.19. C2H2Cl4 + Ca(OH)2 → C2HCl3 + CaCl2 + H2O
Ans. 2C2H2Cl4 + Ca(OH)2 → 2C2HCl3 + CaCl2 + 2H2O
4.20. (NH4)2Cr2O7 → N2 + Cr2O3 + H2O
Ans. (NH4)2Cr2O7 → N2 + Cr2O3 + 4H2O
4.21. Zn3Sb2 + H2O → Zn(OH)2 + SbH3
Ans. Zn3Sb2 + 6H2O → 3Zn(OH)2 + 2SbH3
4.22. HClO4 + P4O10 → H3PO4 + Cl2O7
Ans. 12HClO4 + P4O10 → 4H3PO4 + 6Cl2O7
4.23. C6H5Cl + SiCl4 + Na → (C6H5)4Si + NaCl
Ans. C6H5Cl + SiCl4 + 5Na → (C6H5)4Si + 5NaCl
4.24. Sb2S3 + HCl → H3SbCl6 + H2S
Ans. Sb2S3 + 12HCl → 2H3SbCl6 + 3H2S
4.25. IBr + NH3 → NI3 + NH4Br
Ans. 3IBr + 4NH3 → NI3 + 3NH4Br
4.26. SF4 + H2O → SO2 + HF
Ans. SF4 + 2H2O → SO2 + 4HF
4.27. Na2CO3 + C + N2 → NaCN + CO
Ans. Na2CO3 + 4C + N2 → 2NaCN + 3CO
4.28. K4Fe(CN)6 + H2SO4 + H2O → K2SO4 + FeSO4 + (NH4)2SO4 + CO
Ans. K4Fe(CN)6 + 6H2SO4 + 4H2O → 2K2SO4 + FeSO4 + 3(NH4)2SO4 + 6CO
4.29. Fe(CO)5 + NaOH → Na2Fe(CO)4 + Na2CO3 + H2O
Ans. Fe(CO)5 + 4NaOH → Na2Fe(CO)4 + Na2CO3 + 2H2O
4.30. H3PO4 + (NH4)2MoO4 + HNO3 → (NH4)3PO4 · 12MoO3 + NH4NO3 + H2O
Ans. H3PO4 + 12(NH4)2MoO4 + 21HNO3 → (NH4)3PO4 · 12MoO3 + 21NH4NO3 + 12H2O
4.31. Identify the type of chemical reaction, write the products, and balance the equations.
(a) HCl + Mg(OH)2 →
(b) PbCl2 + K2SO4 →
(c) CH3CH2OH + O2 (excess) →
(d) NaOH + H2C6H6O6 →
(e) Fe + AgNO3 →
Partial answers:
(a) acid-base (neutralization), products are H2O and MgCl2;
(b) double displacement (metathesis), products are KCl and PbSO4;
(c) combustion, products are CO2 and H2O;
(d) acid-base (neutralization), products are Na2C6H6O6 and H2O;
(e) replacement (displacement), products are Fe(NO3)2 and Ag.
4.32. Consider the combustion of amyl alcohol:
(a) How many moles of O2 are required for the complete combustion of 1 mole of amyl alcohol? (b) How many moles of H2O are formed for each mole of O2 consumed? (c) How many grams of CO2 are produced for each mol of amyl alcohol burned? (d) How many grams of CO2 are produced per gram of amyl alcohol burned? (e) How many tons of CO2 are produced per ton of amyl alcohol burned?
Ans. (a) 7.5 mol O2; (b) 0.80 mol H2O; (c) 220 g CO2; (d) 2.49 g CO2; (e) 2.49 tons CO2
4.33. A portable hydrogen generator utilizes the reaction CaH2 + 2H2O → Ca(OH)2 + H2. How many grams of H2 can be produced by a 50-g cartridge of CaH2?
Ans. 4.8 g H2
4.34. Iodine can be made by the reaction 2NaIO3 + 5NaHSO4 → 3NaHSO4 + 2Na2SO4 + H2O + I2. (a) What mass of NaIO3 must be used for each kilogram of iodine produced (b) What mass of NaHSO3/kg I2 is used?
Ans. (a) 1.56 kg NaIO3; (b) 2.05 kg NaHSO3
4.35. Nonane, a component in gasoline, was found to be present at a fire that was suspected to be arson. (a) Write the balance equation for the burning of nonane, C9H20, in air. (b) How many grams of oxygen, O2, are required to burn 500 g nonane? (c) If 32g O2 occupy 22.4 L at 0°C and 1 atm, what volume of oxygen under these conditions is required to burn the nonane?
Ans. (a)C9H20 + 14O2 → 9CO2 + 10H2O; (b) 1750 g O2; (c) 1220 L O2
4.36. The noble gas (Group VIIIA) compound XeF2 can be safely destroyed by treatment with NaOH:
Calculate the mass of oxygen resulting from the reaction of 85.0 g XeO2 with excess NaOH solution.
Ans. 8.03 g O2
4.37. Carbon monoxide, a poisonous gas, is released when nonane burns if there is insufficient oxygen present to produce CO2. (a) Write the balance equation for the burning of nonane producing CO. (b) What mass of CO is released by the burning of 500 g nonane?
Ans. (a) 2C9H20 + 19O2 → 18CO + 20H2O; (b) 938 g CO
4.38. How much iron(III) oxide can be produced from 6.76 g of FeCl3 · 6H2O by the reaction
Ans. 2.00 g Fe2O3
4.39. Zinc blende, ZnS, reacts when strongly when heated in air (2ZnS + 3O2 → 2ZnO + 2SO2). (a) How many pounds of ZnO will be formed when 1 lb zinc blende reacts as above? (b) How many tons of ZnO will be formed from 1 ton ZnS? (c) How many kilograms of ZnO will be formed from 1 kg ZnS?
Ans. (a) 0.835 lb; (b) 0.835 ton; (c) 0.835 kg
4.40. Hydrogen cyanide gas was used as a method of administering the death penalty. Potassium cyanide pellets dropped into an HCl solution yield HCN. (a) Write the balanced reaction producing HCN in this manner. (b) What mass of KCN produces 4 moles HCN gas?
Ans. (a) KCN(aq) + HCl(aq) → HCN(g) + KCl(aq); (b) 260 g KCN
4.41. What mass of HCN can be produced when a 50-g KCN tablet is placed 1 L of solution that contains 6 moles HCl? (Hint: refer to Problem 4.40.)
Ans. 21 g HCN, HCl in a large excess
4.42. During the refining process, silver is to be separated from silver sulfide (argentite) and zinc can be used in the process. Suppose the silver could be liberated directly from the argentite, how many metric tons of zinc would be required to treat 100,000 metric tons of argentite?
Ans. 26,500 metric tons Zn
4.43. Hydrazine, N2H4, can be used as a rocket fuel; the amount of fuel is to be 250,000 kg for a particular space shot. Assuming the products of the reaction with liquid oxygen (LOX) are N2O3 and H2O, what is the mass of N2O3 released during this trip?
Ans. 593,000 kg N2O3
4.44. In a rocket motor fueled with butane, C4H10, how many kilograms of liquid oxygen should be provided with each kilogram of butane to ensure complete combustion?
Ans. 3.58 kg liquid oxygen
4.45. Chloropicrin, CCl3NO2, can be made cheaply for use as an insecticide by the reaction
How much nitromethane, CH3NO2, is needed to form 500 g of chloropicrin?
Ans. 186 g nitromethane
4.46. Carbide lamps were used as a source of light for miners, but they tended to explode if not carefully maintained. Calcium carbide reacts with water to produce acetylene. (a) Write the balanced equation for the reaction. (b) What number of grams C2H2 can be produced from 75.0 g Ca2C?
Ans. (a) Ca2C + 2HCl → C2H2 + CaCl2; (b) 30.5 g C2H2
4.47. (a) CaCO3 is available as an antacid tablet used to neutralize stomach acid, HCl, and knowing that one of the products is CO2, write the balanced equation. (b) How much HCl can be neutralized by a 500-mg tablet available under many market names?
Ans. (a) CaCO3 + 2HCl → CaCl2 + CO2 + H2O; (b) 374 mg HCl neutralized
4.48. (a) CaO + CO2 → CaCO3 is a reaction that provides very pure CaCO3. What mass of CaCO3 can be produced by 1 ton of CaO? (b) What mass of CO2 would be used up?
Ans. (a) 1.8 metric tons CaCO3; (b) 0.78 metric tons CO2
4.49. What mass of the industrial solvent benzene can be produced by the union of 3 acetylene molecules if 100 moles acetylene were to react by 3C2H6 → C6H6?
Ans. 2600 g
4.50. As part of the copper refining process, copper(I) sulfide is oxidized to the reaction Cu2S + O2 → Cu + SO2. SO2 is an environmental problem and cannot be allowed to enter the atmosphere. How many tons of SO2 are released by the reaction of 6 tons Cu2S?
Ans. 2.4 tons SO2
4.51. One of the sources of iron is magnetite, Fe3O4 (FeO and Fe2O3 mixed), which reacts with coke (carbon) to produce iron in the liquid state and carbon monoxide. If we assume that coke is pure carbon, how much is supplied to produce 10,000 metric tons of iron?
Ans. 2075 metric tons coke
4.52. Since CO has been produced in the previous problem and can be used to react with Fe3O4, how much additional magnetite can be reacted to release Fe(l) and CO2?
Ans. 1000 additional metric tons
4.53. Ethyl alcohol (C2H5OH) is made by the fermentation of glucose (C6H12O6). How many metric tons of alcohol can be made from 2.00 metric tons of glucose by the following reaction?
Ans. 1.02 metric tons C2H5OH
4.54. How many kilograms of sulfuric acid can be prepared from 1 kg of cuprite, Cu2S, if each atom of S in Cu2S is converted into 1 molecule of H2SO4?
Ans. 0.616 kg H2SO4
4.55. (a) How much bismuth nitrate, Bi(NO3)3 · 5H2O, would be formed from a solution of 10.4 g bismuth in nitric acid? The reaction is
(b) How much 30.0% nitric acid (containing 30.0% HNO3 by mass) is required to react with 10.4 g Bi?
Ans. (a) 24.1 g; (b) 41.8 g
4.56. One of the reactions used in the petroleum industry for improving the octane rating of fuels is
The two hydrocarbons appearing in this equation are liquids; the hydrogen formed is a gas. What is the percentage reduction in liquid weight accompanying the completion of the above reaction?
Ans. 6.2%
4.57. In the Mond process for purifying nickel, the volatile nickel carbonyl, Ni(CO)4, is produced by the reaction below. How much CO is used up per kilogram of nickel?
Ni + 4CO → Ni(CO)4
Ans. 1.91 kg CO
4.58. When copper is heated with an excess of sulfur, Cu2S is formed. How many grams of Cu2S could be produced if 100 g Cu is heated with 50 g S?
Ans. 125 g Cu2S
4.59. The “thermite” process is of historical interest as a method of welding iron:
Calculate the maximum amount of aluminum that can be mixed with 500 g of iron(III) oxide to form a thermite charge that will yield pure iron.
Ans. 169 g Al
4.60. A mixture of 1 ton of CS2 and 2 tons of Cl2 is passed through a hot reaction tube. The following takes place:
CS2 + 3Cl2 → CCl4 + S2Cl2
(a) How much CCl4, carbon tetrachloride, can be made by complete reaction of the limiting starting material?
(b) Which starting material is in excess, and how much of it remains unreacted?
Ans. (a) 1.45 tons CCl4; (b) 0.28 tons CS2
4.61. A gram (dry weight) of green algae was able to absorb 4.7 × 10–3 mol CO2 per hour by photosynthesis. If the fixed carbon atoms were all stored after photosynthesis as starch, (C6H10O5)n, how long would it take for the algae to double their own weight? (Neglect the increase in photosynthetic rate due to the increasing amount of living matter.)
Ans. 7.9 hours
4.62. Carbon disulfide, CS2, can be made from sulfur dioxide, SO2, a waste product of many industrial processes. How much CS2 can be produced from 450 kg of SO2 with excess coke, if the SO2 conversion is 82% efficient? The overall reaction is
5C + 2SO2 → CS2 + 4CO
Ans. 219 kg
4.63. Methane, CH4, has been proposed as a motor fuel by governmental officials and some popular movies. (a) How much water would be produced by the complete burning of 3500 g CH4 by CH4 + 2O2 → CO2 + 2H2O? (b) Actually, there is a loss of horsepower when using methane as a fuel for internal combustion engines—propose an explanation.
Ans. (a) 3900 g H2O; (b) Methane has only one carbon (4 hydrogens), but gasoline is a mixture of hydrogen-carbon compounds containing much more carbon (4–12 carbons and lots of hydrogen) per mole. The carbon and hydrogen are oxidized and give off the heat necessary for the functioning of internal combustion engines. Gasoline simply provides more heat per mole.
4.64. Ethanol, C2H5OH, is the component in alcoholic beverages that produces the neurological disturbances associated with over-use. Ethanol is metabolized to CO2 and H2O over time (approximately 20 g/h) by reaction with O2. (a) Write the balanced equation for the metabolism of ethanol. (b) How much oxygen is required to metabolize one hour’s consumption? (c) The equation is the same one for the burning of alcohol as a fuel when mixed with gasoline (gasohol). How much oxygen is needed to burn 1 L of alcohol (density: 0.789 g/mL)?
Ans. (a)2C2H5OH + 7O2 → 4CO2 + 6H2O; (b) 489 oz O2 (1380 g); (c) 1920 g O2
4.65. Silicate ores can be solubilized by fusion with sodium carbonate. A simplified equation for what occurs is
Calculate the minimum weight of Na2CO3 required to dissolve a 0.500-g sample of an ore which analyzes 19.1% silica (SiO2).
Ans. 0.337 g (Note: A large excess is normally used—about 3 g.)
4.66. The chemical formula of the chelating agent Versene is C2H4N2(C2H2O2Na)4. If each mole of this compound could bind 1 mole of Ca2+, what would be the rating of pure Versene, expressed as mg CaCO3 bound per gram of chelating agent? Here, the Ca2+ is expressed in terms of the amount of CaCO3 it could form.
Ans. 264 mg CaCO3 per g
4.67. When calcium carbide, CaC2, is made in an electric furnace by the reaction below, the crude product is typically 85% CaC2 and 15% unreacted CaO. How much CaO is to be added to the furnace charge for each 50 tons (a) of CaC2 produced? (b) of crude product?
CaO + 3C → CaC2 + CO
Ans. (a) 53 tons CaO; (b) 45 tons CaO
4.68. The plastics industry uses large amounts of phthalic anhydride, C8H4O3, made by the controlled oxidation of naphthalene:
Since some of the naphthalene is oxidized to other products, only 70% of the maximum yield predicted by the above equation is actually obtained. How much phthalic anhydride would be produced in practice by the oxidation of 100 lb C10H8?
Ans. 81 lb
4.69. The empirical formula of a commercial ion-exchange resin is C8H7SO3Na. The resin can be used to soften water according to the reaction provided. Expressed in moles Ca2+ taken up per gram resin used, what would be the maximum uptake of Ca2+?
Ans. 0.0024 mol Ca2+/g resin
4.70. The insecticide Chlordane is made by a two-step process:
(a) Calculate how much Chlordane can be made from 500 g of C5Cl6. (b) What is the percentage of chlorine in Chlordane?
Ans. (a) 751 g; (b) 69.2%
4.71. Commercial sodium “hydrosulfite” is 90% pure Na2S2O4. How much commercial product could be made using 100 tons of zinc with a sufficient supply of the other reactants? The reactions are
Ans. 296 tons
4.72. Fluorocarbon polymers can be made by fluorinating polyethylene according to the reaction
where n is a large integer. The CoF3 can be regenerated by the reaction
(a) If the HF formed in the first reaction cannot be reused, how many kilograms of fluorine are consumed per kilogram of fluorocarbon produced, (CF)n? (b) If the HF can be recovered and electrolyzed to hydrogen and fluorine, and if this fluorine is used for regenerating CoF3, what is the net consumption of fluorine per kilogram of fluorocarbon?
Ans. (a) 1.52 kg; (b) 0.76 kg
4.73. A process designed to remove organic sulfur from coal prior to combustion involves the following reactions:
In the processing of 100 metric tons of coal having a 1.0% sulfur content, how much limestone (CaCO3) must be decomposed to provide enough Ca(OH)2 to regenerate the NaOH used in the original leaching step?
Ans. 3.12 metric tons
4.74. Silver may be removed from solutions of its salts by reaction with metallic zinc:
A 50-g piece of zinc was thrown into a 100-L vat containing 3.5 g Ag+/L. (a) Which reactant was completely consumed? (b) How much of the other substance remained?
Ans. (a) zinc; (b) 1.9 g Ag+/L
4.75. The following reaction proceeds until the limiting substance is completely consumed:
2Al + 3MnO → Al2O3 + 3Mn
A mixture containing 100 g Al and 200 g MnO was heated to initiate the reaction. Which reactant remained in excess and what was the weight of that substance?
Ans. Al, 49 g
4.76. A mixture of NaHCO3 and Na2CO3 weighed 1.0235 g. The dissolved mixture was reacted with excess Ba(OH)2 to form 2.1028 g BaCO3 by the reactions
What was the percentage NaHCO3 in the original mixture?
Ans. 39.51% NaHCO3
4.77. A mixture of NaCl and NaBr weighed 3.5084 g. This mixture was dissolved and treated with enough AgNO3 to precipitate all the chloride and bromide as AgCl and AgBr. The washed precipitate was treated with KCN to increase solubility of the silver, then the solution was electrolyzed. The process is
After the final step was completed, 5.5028 g pure silver were collected. What was the composition of the initial mixture?
Ans. 65.23% NaCl, 34.77% NaBr