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)

Answer 1

First, write a correctly balanced equation for the reaction taking place:

2Al(s) + 6HCl(aq) ===> 2AlCl₃(aq) + 3H₂(g) ... balanced equation

moles H₂(g) required:

PV = nRT and solve for n (moles)

n = PV/RT = (1.11 atm)(4.04 L) / (0.0821 Latm/Kmol)(300K) ... note to change T to Kelvin

n = 0.1821 moles H₂

moles Al(s) required to generate 0.1821 moles H₂(g):

0.1821 moles H₂ x 2 mol Al / 3 moles H₂ = 0.121 moles Al (to 3 significant figures)