Question

# Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water...

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?

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 %

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