Zn(s) + 2 HCl(aq) → H2(g) + ZnCl₂(aq) When 25.0 g of Zn reacts, how many L of H₂ gas are formed at 25 °C and a pressure of 854 mmHg?

Question

Quizplus · Accepted Answer

First, we need to use stoichiometry to determine the moles of H2 produced:
25.0 g Zn * (1 mol Zn/65.38 g Zn) * (1 mol H2/1 mol Zn) = 0.383 mol H2
Next, we can use the ideal gas law to determine the volume of gas produced:
PV = nRT
V = nRT/P
V = (0.383 mol)(0.0821 L·atm/mol·K)(298 K)/(854 mmHg)(1 atm/760 mmHg)
V = 8.32 L 
Therefore, the correct answer is (C) 8.32 L.