Stoichiometry Examples - Explanations and Answers

1. In an aqueous solution, barium chloride and potassium chromate react to form solid barium chromate, with potassium chloride being left behind in solution. Using the balanced equation, calculate how much barium chromate you would expect to produce if you mixed 1.534 grams of solid barium chloride with an excess amount of 0.25 molar (0.25 M) potassium chromate solution.

For this problem you need to balance the reaction; the only coefficient needed is the two in front of the potassium chloride. The potassium chromate is excess, so the molarity is not important here, but it's in the problem anyway, so you have to be able to figure out that you don't need it. This is a straight stoichiometry problem, where you'll use molar mass to convert grams to moles of barium chloride, then use the coefficients to convert moles of barium chloride to moles of barium chromate, then use the molar mass of barium chromate to get the grams. Don't forget to write out your molar mass calculations for barium chloride and barium chromate. Also, don't forget to show the mole-to-mole conversion, even though both coefficients in the conversion are one. Answer: 1.866 grams of barium chromate.

2. Calculate the percentage yield if your actual yield of barium chromate from problem #1 was 0.893 grams.

Percent yield is actual yield over theoretical yield, times 100 (to convert the ratio to percent). You calculated the theoretical yield in the previous problem; this is the amount you'd expect to get, based on the starting amounts of each reactant. The actual yield is given as 0.893 grams; this is the amount actually obtained in the lab. Answer: 47.8%.

3. Silver chloride can be precipitated by the reaction of silver nitrate and sodium chloride. The other product, which remains in solution, is sodium nitrate. Suppose you have 13.7 grams of a material containing an unknown amount of silver nitrate. When you dissolve this material in water and react it with an excess amount of sodium chloride, you produce 8.4 grams of silver chloride. Assuming that all of the silver nitrate in the mixture reacted to form silver chloride, what was the mass percentage of silver nitrate in the material?

This problem is similar to one we did in lab. You need to first write out the reaction and balance it. All coefficients are one. Now, the 13.7 grams is the mass of a mixture containing silver nitrate, so when you figure out the percent of silver nitrate in the mixture, this will be in the denominator of the percent yield fraction. You can calculate the amount of silver nitrate in the mixture (the numerator of the percent yield fraction) by working backwards from the amount of silver chloride obtained in the reaction. This is another simple gram-to-mole-to-mole-to-gram stoichiometry conversion. Answer: 73%.

4. In the booster rocket for the space shuttle, aluminum metal reacts with ammonium perchlorate to form aluminum oxide, aluminum chloride, nitrogen monoxide, and water (plus lots of energy). How much ammonium perchlorate is required to react with each 1.000 kilograms of aluminum metal?

This is a pretty complex reaction. It has two reactants and four products. You need to make sure you read the problem carefully enough to get it right. Then balance it. The coefficients on both reactants are three; the coefficients on water and nitrogen monoxide are six and three, respectively; the other coefficients are one. After you balance the reaction and convert kilograms to grams, it's another simple stoichiometric conversion from grams of aluminum to moles of aluminum to moles of ammonium perchlorate. Since the question doesn't really specify units for the answer, you can just leave it at moles (since this clearly expresses "how much" of the ammonium perchlorate is needed). Answer: 37.06 moles of NH₄ClO₄.

5. A bottling plant has 121,500 bottles, each with a capacity of 355 milliliters. It also has 125,000 caps and 53,575 liters of beverage. How many bottles can be completely filled with the beverage and capped?

This is a limiting ingredient (reactant) problem. You can tell because the problem gives the amounts of each reactant: bottles, caps, and beverage. The "reaction" is 1 bottle + 1 cap + 355 ml beverage yields 1 full and capped bottle. Use the reaction to figure out how many full and capped bottles can be made from 121,500 bottles, assuming an unlimited amount of beverage and an unlimited number of caps; then use it again to figure out how many full and capped bottles can be made from 125,000 caps and unlimited bottles and beverage; then use it again to figure out how many full and capped bottles can be made from 53,575 liters of beverage and unlimited caps and bottles. The smallest answer of the three is the right one. Don't forget to convert liters to milliliters. Answer: 121,500 full and capped bottles of soda.

6. Silicon carbide (SiC) is an abrasive formed by the reaction of silicon dioxide with elemental carbon at high temperatures. The byproduct of this reaction is carbon monoxide. How many grams of silicon carbide can be formed from the reaction of 2.50 grams of silicon dioxide with 2.50 grams of carbon? What is the limiting reactant? How much of the excess reactant is leftover?

This is another limiting reactant problem. Again, you can tell because you know the starting amounts of both reactants. First, set up the reaction and balance it. The coefficient on carbon is 3; on carbon monoxide, it's 2; the other coefficients are 1. Start with the mass of one reactant, silicon dioxide or carbon, convert to moles, then to moles of silicon carbide and grams of silicon carbide. Then do the same for the other reactant. You should now have two numbers for the mass of silicon carbide produced. The smaller one is the answer. Answer: 1.67 grams of silicon carbide. Silicon dioxide is the limiting reactant.

To calculate the amount of carbon leftover, you need to work backwards from the amount of silicon carbide actually produced (i.e., your answer). Start with the 1.67 grams of silicon carbide, convert to moles of silicon carbide, then to moles of carbon, and finally to grams of carbon. This is the amount of carbon CONSUMED, not the amount leftover. To calculate the amount leftover, subtract the amount consumed from the amount originally available. Answer: 1.00 grams of carbon leftover.

7. Glucose (C₆H₁₂O₆) burns in oxygen to produce carbon dioxide and water. How many grams of oxygen are required to burn 10.0 grams of glucose? How many grams of water are produced from this reaction?

This is a two-part stoichiometry problem. First, write out the reaction and balance it. Remember that oxygen is a diatomic gas in its natural state. Start with the mass of glucose consumed (burned), convert to moles of glucose, then to moles of oxygen, then to grams of oxygen. Remember to show your molar mass calculations. Answer: 10.7 grams of oxygen.

For the second part, start with the mass of glucose consumed, convert to moles of glucose, then to moles of water, then to grams of water. Answer: 6.00 grams of water.

8. If 10.0 grams of glucose burn in 10.0 grams of oxygen, how many grams of carbon dioxide would be produced? What was the limiting reactant? How many grams of the excess reactant would be leftover?

Another limiting reactant problem. Answer: 13.8 grams of carbon dioxide. Oxygen was the limiting reactant.

To find the mass of glucose leftover, you need to calculate the mass of glucose that was consumed. Start with the amount of carbon dioxide actually produced (13.8 grams), convert to moles of carbon dioxide, then to moles of glucose, then to grams of glucose. Subtract the mass of glucose consumed from the mass available (10.0 grams) to get the mass of glucose leftover. Answer: 0.6 grams of glucose leftover.

9. Gastric fluid contains about 3.0 grams of hydrogen chloride per liter. If a person produces about 2.5 liters of gastric fluid per day, how many antacid tablets, each containing 400. milligrams of aluminum hydroxide, are needed to neutralize all of the hydrogen chloride produced in one day? The two products of the neutralization reaction are aluminum chloride and water.

This is actually a pretty simple stoichoimetry problem, as long as you keep track of the unit conversions. Your goal is to calculate the number of tablets needed to neutralize one day of acid-producing stomach activity. Start with the one day. Assume it's exact, because there are no digits written in the problem. (If it was 1.0 days, that would be two sig figs, for example.) Convert days to liters of acid, then to grams of acid, using the conversion factors given. Use the molar mass of hydrochloric acid to convert from grams to moles of acid. Then use the balanced reaction to convert to moles of aluminum hydroxide. Use the molar mass of aluminum hydroxide to get the grams, then convert from grams to number of tablets using the conversion factor given. Answer: 14 tablets. (Normally, you'd round down from 13.38 tablets, but in this case, you want to neutralize all of the acid, so if you round down, you won't quite get all of it. So, since I'm limited to two sig figs by the 3.0 grams of acid, I rounded up to 14.)

10. Acetylene (C₂H₂) can be manufactured by the reaction of calcium carbide (The formula is not what you'd expect. It is CaC₂.) with water. The other product of the reaction is calcium hydroxide. When 44.5 grams of commercial grade (impure) calcium carbide are reacted with an excess amount of water, 0.540 grams of acetylene are produced. Assuming that all of the calcium carbide reacted to form acetylene, what is the percent of calcium carbide in the commercial grade material?

This is another mixture problem, as "impure" calcium chloride is a mixture of calcium carbide and something else. The equation is pretty easy to balance: all coefficients are one except for water's, which is two. You need to work backwards from the mass of acetylene actually produced to the moles of acetylene to the moles of calcium carbide to the mass of calcium carbide. The percent of calcium carbide is the mass of calcium carbide that reacted to form the acetylene, divided by the mass of the mixture, times 100. Answer: 2.99%.