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

Question

Quizplus · Accepted Answer

Using stoichiometry, 262 g of CH₄ is 16.32 moles. From the balanced equation, 1 mole of CH₄ produces 3 moles of H₂. Therefore, 16.32 moles of CH₄ will produce 48.96 moles of H₂. Using the ideal gas law (PV=nRT), with R=0.0821 L·atm/mol·K, T=423 K, and P=0.862 atm, the volume V = (nRT)/P = (48.96 * 0.0821 * 423) / 0.862 = 1970 L.

Hydrogen Gas (2 →\to→ CO(g)+ 3 H2(g) If 262 G CH4 React with Excess

Hydrogen Gas (2 $\to$ CO(g)+ 3 H₂(g)
If 262 G CH₄ React with Excess