In the reaction between Al and HCl, if 1.32 g Al were reacted with excess hydrochloric acid, how many mL of Hydrogen gas could theoretically be emitted at STP?

Question

In the reaction between Al and HCl, if 1.32 g Al were reacted with excess hydrochloric acid, how many mL of Hydrogen gas could theoretically be emitted at STP?

the balanced equation should be:

2 Al + 6 HCl —> 2 AlCl3 + 3 H2

Answer 1

2 Al + 6 HCl —> 2 AlCl₃ + 3 H₂ ... balanced equation

NOTE: the mole ratio of H2 gas to Al is 3:2 (3 mol H2 produced from 2 mol Al)

moles Al used = 1.32 g Al x 1 mol Al/26.98 g = 0.0489 moles Al

moles H2 produced = 0.0489 mol Al x 3 mol H2/2 mol Al = 0.0734 moles H2 gas produced

At STP, 1 mole of an ideal gas = 22.4 L, so for 0.0734 moles of H2 gas, we have the following relationship:

0.0734 moles H2 x 22.4 L/mole = 1.64 lites of H2 gas

To convert to mls, just multiply by 1000: 1.64 L x 1000 ml/L = 1640 mls of H2 gas