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%

h let me think

idk what the question is about