When 14.0 g of zinc metal reacts with excess HCl, how many litres of H₂ gas are produced at STP? Zn(s)+ 2HCl(aq)→ ZnCl₂(aq)+ H₂(g)

Question

Quizplus · Accepted Answer

First, we need to balance the equation:
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Using stoichiometry, we can calculate the moles of H2 produced:
n(H2) = n(Zn) = m(Zn) / MM(Zn) = 14.0 g / 65.38 g/mol = 0.2142 mol
Since the reaction takes place at STP, 1 mole of gas occupies 22.4 L:
V(H2) = n(H2) x 22.4 L/mol = 0.2142 mol x 22.4 L/mol = 4.8 L
Rounding to two significant figures, the answer is 4.86 L. Therefore, the best choice is C.

When 140 G of Zinc Metal Reacts with Excess HCl, How