How many moles of aluminum metal are required to produce 4.04 l of hydrogen gas at 1.11 atm and 27 °c by reaction with hcl?
2 al(s) + 6 hcl(aq) → 2 alcl3 (aq) + 3 h2(g)

0.121 moles of aluminum metal are required to produce 4.04 L of hydrogen gas at 1.11 atm and 27 °C by reaction with HCl

Explanation:

This is the reaction:

2 Al(s) + 6 HCl(aq) → 2 AlCl₃ (aq) + 3 H₂(g)

To make 3 moles of H₂, we need 2 moles of Al.

By conditions given, we will find out how many moles of H₂ do we have.

Let's use the Ideal Gas Law

P. V = n . R . T

1.11 atm . 4.04L = n . 0.082 L.atm/mol.K . 300K

(1.11 atm . 4.04L) / (0.082 mol.K/L.atm . 300K) = n

0.182 mol = n

So the rule of three will be:

If 3 moles of H₂ came from 2 moles of Al

0.182 moles of H₂ will come from x

(0.182 .2) / 3 = 0.121 moles