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, 25.5 L of methane gas (measured at a pressure of 734 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 26.4 L of hydrogen gas measured at STP.

Let us assume that all gases are ideal. So we can use the formula:

PV = nRT

The reaction is:

CH4 (g) + H2O (g) → CO (g) + 3H2 (g)

First we determine what the limiting reactant is. This can be done by calculating for the number of moles (n) for each reactant.

For CH4:

nCH4 = (734 torr) (25.5 L) / (62.36367 L torr / mol K) (298.15 K)

nCH4 = 1 mol

For H2O:

nH2O = (704 torr) (22.6 L) / (62.36367 L torr / mol K) (398.15 K)

nH2O = 0.64 mol

Therefore H2O is the limiting reactant therefore the theoretical moles of H2 produced is:

nH2 (theo) = 0.64 mol * (3 mol H2 / 1 mol H2O) = 1.92 mol

The actual moles of H2 is:

nH2 (actual) = (750 torr) (26.4 L) / (62.36367 L torr / mol K) (273.15 K) = 1.16 mol

Therefore the yield is:

% yield = 1.16 / 1.92 * 100%

% yield = 60.42%