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 730 torr and a temperature of 25 ∘C) is mixed with 22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 ∘C). The reaction produces 26.2 L of hydrogen gas measured at STP.

What is the percent yield of the reaction?

Answer: 60.8%

Answer 1

CH4(g) + H2O(g) → CO(g)+ 3H2(g)

Using ideal gas law: PV = nRT

calculating no. of moles of methane:

n = PV/RT

=(730/760 atm)(26.5 L)/(0.0821 L.atm/K.mol)(298.15 K) = 1.04 mol

calculating no. of moles of warer vapor:

n = PV/RT

=(704/760 atm)(22.6 L)/(0.0821 L.atm/K.mol)(398.15 K) = 0.64 mol

calculating no. of moles of hydrogen gas:

molar volume of H₂ at STP = 22.4 L/mol

no. of moles of H₂ in 26.2 L = 26.2/22.4 = 1.16 mol

From the balanced equation,

1 mol of methane needs 1 mole of water vapor to produce 3 moles of hydrogen gas.

From the calculations above we find that-

no. of moles of methane = 1.04

no. of moles of water vapor = 0.64

Hence, limiting reagent is water vapor.

moles of hydrogen produced = 3 x 0.64 = 1.92 mol

Actual yield = 1.16 mol

Hence, percent yield = (actual yield)/(theoretical yield)x100

=(1.16/1.92)x100 = 60.4 %