When 14.0 g of zinc metal reacts with excess HCl,how many litres of H₂ gas are produced at STP? $$\mathrm { Zn } ( s ) + 2 \mathrm { HCl } ( a q ) \rightarrow \mathrm { ZnCl } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } ( g )$$ A)0.208 L B)0.416 L C)4.86 L D)9.60 L

Question

Quizplus · Accepted Answer

The balanced chemical equation shows that 1 mole of Zn yields 1 mole of H2.

First, we need to find the number of moles of Zn:

m(Zn) = 14.0g / 65.38 g/mol = 0.214 mol

Since 1 mole of Zn yields 1 mole of H2, we have:

n(H2) = 0.214 mol

At STP, one mole of gas occupies 22.4 L, so:

V(H2) = n(H2) x 22.4 L/mol = 0.214 mol x 22.4 L/mol = 4.79 L

Therefore, the answer is C) 4.86 L (rounded to two significant figures).

When 140 G of Zinc Metal Reacts with Excess HCl,how Many