Question

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, 26.5 L of methane gas (measured at a pressure of 736 torr and a temperature of 25 ∘C) is mixed with 22.6 L of water vapor (measured at a pressure of 700 torr and a temperature of 125 ∘C). The reaction produces 26.0 L of hydrogen gas measured at STP.

What is the percent yield of the reaction?

Answer 1

We need to first convert torr to atm,

since 1 atm = 760 torr

736 torr = 0.97 atm
700 torr = 0.92 atm

and

25 °C = 298 K
125 °C = 398 K

By PV = nRT, the number on moles are

PV/RT = n

the mols of CH4

(0.97 atm * 26.5 L)/(0.082 (L atm) /(mol K) * 298) = 1.05 mol CH4

(0.92 atm * 22.6 L)/(0.082 (L atm) /(mol K) * 398) = 0.64 mol H2O

By the stoichiometric equation

CH4(g) + H2O(g) ⇄ CO(g) + 3H2(g)

for every mol of CH4 number of mols of H2 produced = 3

3 * 0.64 mol H2O = 1.91 mols H2

Again by PV = nRT,  

V = nRT / P

At STP the pressure is 1 atm, the temperature is 273 K

(1.91 mol * 0.082 (L atm) /(mol K) * 273) / 1 atm = 42.8 L

This is the theoretical yield, the actual yield was 26.0 L

so percent yield = theoretical yield/actual yield *100
26.0 / 42.8 * 100 = 60.75 %