What volume of hydrogen (STP) is produced by dissolving 1.1 moles of aluminum in sulfu- ric acid? 2 Al(s) + 3 H2SO4 → Al2(SO4)3 + 3H2

Answer in units of L.​

37 L

Explanation:

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

2 moles of Al react to 3 moles of sulfuric acid, in order to produce 1 mol of aluminum sulfate and 3 moles of hydrogen.

So ratio between aluminum and hydrogen is 2:3. We assume the acid is the limiting reactant.

2 moles of Al can produce 3 moles of hydrogen

Then 1.1 moles of Al, will produce (3 . 1 .1)/2 = 1.65 moles

If 1 mol of any gas, occupies 22.4L at STP

1.65 moles of H₂ may ocuppy (1.65 . 22.4) / 1 = 36.96 L ≅ 37L

If we had applied Ideal Gases Law → P . V = n . R  .T

At STP. we have 1 atm and 273 K

V = (n . R . T) / P → (1.65 . 0.082 . 273) / 1 = 36.9 L

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