Problem: 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) + 3 H2 (g)In a particular reaction, 26.0 L of methane gas (measured at a pressure of 734 torr and a temperature of 25 oC) is mixed with 22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 oC). The reaction produces 26.6 L of hydrogen gas measured at STP.What is the percent yield of the reaction?

CH₄(g) + H₂O(g) → CO(g) + 3 H₂ (g)

Calculate how much H₂(g) is produced from each reactant:

$$\begin{gathered} \boxed{\mathbf{PV}\mathbf{=}\mathbf{nRT}} \\ \mathbf{n}\mathbf{=}\frac{\mathbf{PV}}{\mathbf{RT}} \end{gathered}$$

H₂(g) from CH₄(g):

$\mathbf{P}\mathbf{=}\mathbf{734}\mathbf{ }\cancel{\mathbf{torr}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmHg}}{\mathbf{760}\mathbf{ }\cancel{\mathbf{torr}}}$

P = 0.9658 atm

V = 26.0 L
T = 25°C + 273.15 = 298.15 K

${\mathbf{n}}_{{\mathbf{CH}}_{\mathbf{4}}}\mathbf{=}\frac{(0.9658 \cancel{\mathrm{atm}})(26.0 \cancel{L})}{(0.08206{\displaystyle \frac{\cancel{L}·\cancel{\mathrm{atm}}}{\mathrm{mol}·\cancel{K}}})(298.15 \cancel{K})}$

n_CH4 = 1.0263 mol

$\mathbf{moles}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0263}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{CH}}_{\mathbf{4}}}\mathbf{\times }\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{CH}}_{\mathbf{4}}}}$

moles H₂ = 3.079 mol

H₂(g) from H₂O(g):

22.6 L of water vapor (measured at a pressure of 704 torr and a temperature of 125 ^oC

$\mathbf{P}\mathbf{=}\mathbf{704}\mathbf{ }\cancel{\mathbf{torr}}\mathbf{\times }\frac{\mathbf{1}\mathbf{ }\mathbf{mmHg}}{\mathbf{760}\mathbf{ }\cancel{\mathbf{torr}}}$

P = 0.9263 atm

V = 22.6 L
T = 125°C + 273.15 = 398.15 K

${\mathbf{n}}_{{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{=}\frac{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{9263}\mathbf{ }\cancel{\mathbf{atm}}\mathbf{)}\mathbf{(}\mathbf{22}\mathbf{.}\mathbf{6}\mathbf{ }\cancel{\mathbf{L}}\mathbf{)}}{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{08206}{\displaystyle \frac{\cancel{L}·\cancel{\mathrm{atm}}}{\mathrm{mol}·\cancel{K}}}\mathbf{)}\mathbf{(}\mathbf{398}\mathbf{.}\mathbf{15}\mathbf{ }\cancel{\mathbf{K}}\mathbf{)}}$

n_H2O = 0.6407 mol

$\mathbf{moles}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{6407}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}\mathbf{\times }\frac{\mathbf{3}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}}}$

moles H₂ = 1.922 mol

H₂O(g): produces a smaller amount of H₂(g) → H₂O(g) is the limiting reactant

theoretical yield = 1.922 mol H₂

Calculate the theoretical yield of H₂(g) (volume):