Answer:
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
