If 20.0 grams of Al is placed into a solution containg

Chemistry

1 answer:

meriva2 years ago

5 0

Answer:

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

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