Question

What volume of H2 could be produced at a temperature of 22 oC and a pressure of 751 torr by the reaction of 2.24 g of aluminum with hydrochloric acid?

____ L H2

Answer 1

Given T = 22 ⁰C = 22 ⁰ C + 273.15 = 295.15 K

Pressure in torr = 751 torr

Mass of Al = 2.24 g

Solution
First we convert pressure into atm

Pressure in atm

= 751 torr x 1 atm / 760 torr

= 0.98816 atm

Reaction

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

Calculation of moles of H2

Moles of H2 = Moles of Al x 3 moles of H2 / 2 mol Al

= (Mass of Al / molar mass of Al ) x 3 moles of H2 / 2 mol Al

= ( 2.24 g Al / 26.9816 g per mol ) x 3 moles of H2 / 2 mol Al

= 0.08302 mol Al x 3 moles of H2 / 2 mol Al

= 0.124529 mol H2

Calculation of volume of H2

We use Ideal gas law

pV = nRT

p is pressure in atm , V is volume in L , n is number of moles, R is gas constant, T is temperature in K

V = nRT / p

R = 0.08206 L atm / (mol K )

= 0.124529 mol x ( 0.08206 L atm / (mol K ) ) x 295.15 K / 0.98816 atm

= 3.052 L

So the volume of H2 produced = 3.05 L