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 determine the moles of Zn present in the reaction:

25.0 g Zn x (1 mol Zn/65.38 g Zn) = 0.383 mol Zn

According to the balanced equation, 1 mole of Zn produces 1 mole of H₂ gas. Therefore, we have 0.383 mol of H₂ gas.

Next, we can use the ideal gas law to calculate the volume of H₂ gas at the given temperature and pressure:

PV = nRT

V = (nRT)/P

V = (0.383 mol)(0.08206 L atm/mol K)(298 K)/(854 mmHg)(1 atm/760 mmHg)

V = 8.32 L

Therefore, the answer is (C) 8.32 L.