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 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%

Homework Answers

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 %

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