Question

(10) Oxygen gas reacts with powdered aluminum according to the following reaction: 4Al(s)+3O2(g)→2Al2O3(s). Part A What volume of O2 gas (in L), measured at 778 mmHg and 19 ∘C, is required to completely react with 53.3 g of Al? Part B Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the following equation: CH4(g)+H2O(g)→CO(g)+3H2(g) In a particular reaction, 26.0 L of methane gas (measured at a pressure of 732 torr and a temperature of 25 ∘C) is mixed with 22.8 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 ∘C). The reaction produces 27.0 L of hydrogen gas measured at STP. What is the percent yield of the reaction?

Answer 1

(10)

Part A) moles of Al = 53.3 g/27 g/mol = 1.974 mol

moles of O2 required = 1.974 mol x 3/4 = 1.4805 mol

Volume of O2 required = nRT/P = 1.4805 x 0.08205 x (273 + 19)/(778/760) = 34.65 L

Part B) Taking pressure in atm, Temperature in K

moles of CH4 = PV/RT = 0.963 x 26/0.08205 x (273 + 25) = 1.024 mol

moles of H2O = 0.926 x 22.8/0.08205 x (273 + 125) = 0.646 mol

Theoretical yield of H2 at STP = 3 x 0.626 mol x 22.4 L = 43.41 L

Percent yield = 27 x 100/43.41 = 62.20%