If 20.0 grams of Al is placed into a solution containg

115grams of H2SO4, how many grams of

hydrogengas could be produced?

m H2 (g) = 2.241 g H2 (g)

Explanation:

2Al (s) + 3H2SO4 (aq) → Al2 (SO4) 3 (aq) + 3 H2 (g)

limit reagent:

∴ Mw Al = 26.982 g/mol

∴ Mw H2SO4 = 98.0785 g/mol

⇒ n Al = (20 g Al) * (mol/26.982 g) = 0.7412 mol Al

⇒ n H2SO4 = (115 g H2SO4) * (mol/98.0785 g) = 1.173 mol H2SO4

⇒ n H2 = (0.7412 mol Al) * (3 mol H2 / 2 mol Al) = 1.112 mol H2

∴ Mw H2 = 2.016 g/mol

⇒ g H2 = (1.112 mol H2) * (2.016 g/mol) = 2.241 g H2