Hydrogen gas (2.02 g/mol) can be produced from the reaction of methane (16.05 g/mol) and water vapor (18.02 g/mol) according to the following reaction equation: CH₄ (g) +H₂ O(g) $\rightarrow$ CO(g) + 3H₂ (g)
If 262 g of CH₄ reacts with excess H₂O at 423 K and 0.862 atm, what volume of H₂ gas will form?

Question

Hydrogen gas (2.02 g/mol) can be produced from the reaction of methane (16.05 g/mol) and water vapor (18.02 g/mol) according to the following reaction equation: CH₄ (g) +H₂ O(g)

\rightarrow

CO(g) + 3H₂ (g)
If 262 g of CH₄ reacts with excess H₂O at 423 K and 0.862 atm, what volume of H₂ gas will form?

Quizplus · Accepted Answer

First, calculate the moles of CH4 (methane) using its molar mass: 262 g / 16.05 g/mol = 16.32 moles. The reaction produces 3 moles of H2 for every mole of CH4, so 16.32 moles of CH4 will produce 16.32 * 3 = 48.96 moles of H2. Using the ideal gas law (PV=nRT), where P = 0.862 atm, n = 48.96 moles, R = 0.0821 L·atm/(mol·K), and T = 423 K, solve for V: V = nRT/P = (48.96 moles * 0.0821 L·atm/(mol·K) * 423 K) / 0.862 atm = 1970 L.

Hydrogen Gas (2 →\rightarrow→ CO(g) + 3H2 (G) If 262 G of CH4 Reacts

Hydrogen Gas (2 $\rightarrow$ CO(g) + 3H₂ (G)
If 262 G of CH₄ Reacts