Honors Chemistry Additional Chapter 12 Problems
These are difficult, multi-step problems. Use the “Brick Wall” method of problem solving to solve these problems. The answers are on the second page of this document.
1. Assume that the human body requires daily energy that comes from metabolizing 816 g of sucrose, C12H22O11, using the following equation:
C12H22O11 + 12 O2 → 12 CO2 + 11 H2O + energy
a) How many dm3 (liters) of pure oxygen (at STP) are consumed by a human being in 24 hours?
b) Suppose a thief accidently gets locked in a 35.5 m3, air-tight bank vault during a robbery. The vault cannot be opened for 24 hours. If the air in the vault contains 20% oxygen by volume, can the thief light a 224 g candle (wax, C25H52, undergoes a combustion reaction when burned), let it burn to completion, and still survive to be arrested when the vault opens?
2. A student has a mixture of KClO3, K2CO3, and KCl. She heats 50.0 g of the mixture and determines that 5.00 g of oxygen and 7.00 g of carbon dioxide are produced by these reactions:
2KClO3 → 2KCl + 3 O2
K2CO3 → K2O + CO2
KCl is not affected by the heat. What is the percent composition of the original mixture?
Answers:
1.a) 641 dm3
1.b) Candle burns – uses up 542 L dm3 (L) of oxygen.
542 L + 641 L (amount used by thief to breather) = 1183 L needed by thief and candle.
35.5 m3 bank vault (35, 000 dm3) filled with air (20% oxygen) = 7100 dm3 (L) of oxygen available for thief and candle. The thief lives.
3. 5.00 g of oxygen given off means 12.8 grams of potassium chlorate decomposed
7.00g of carbon dioxide released means 22.0 g of potassium carbonate decomposed.
50.0 – 12.8 – 22.0 = 15.2 g of KCl since that is not affected by the heat
% potassium chlorate = 25.6%
% potassium carbonate = 44.0 %
% potassium chloride = 30.4%