In the reaction 2Al + 3H2SO4 ----------→ Al2(SO4)3 + 3H2 a) Find the amount of Al used to liberate 22.1 lit. of H2 at STP. b) When 5.4 gm of Al and 9.8 gm of H2SO4 are allowed to react what is the volume of H2 evolved at STP?
Answers
Given : In the reaction, 2Al + 3H₂SO₄ ----→ Al₂(SO₄)₃ + 3H₂
To find : (a) The amount of Al used to liberate 22.1 litre of H₂ gas at STP.
(b) When 5.4 gm of Al and 9.8 gm of H2SO4 are allowed to react what is the volume of H2 evolved at STP
solution : (a) given volume of H₂ = 22.1 litre
⇒no of mole of H₂ = 22.1/22.4 = 0.9866 mol
two moles of Al are required to liberate 3 moles of H₂ gas.
⇒no of moles of Al is required to liberate 0.9866 mol = 2/3 × 0.9866 = 0.657 mol
mass of Al = 0.657mol × 27 g/mol = 17.74 g
(b) two moles of Al react with 3 moles of sulphuric acid.
⇒2 × 27g = 54g of Al reacts with 3 × 98 = 294 g of sulphuric acid.
⇒5.4 g of Al reacts with 294/54 × 5.4 = 29.4g of sulphuric acid. but only 9.8g of sulphuric acid is given. so, H₂SO₄ is limiting reagent.
no of moles of Sulphuric acid = given mass/molecular mass = 9.8/98 = 0.1
now 3 moles of Sulphuric acid give 3 moles of hydrogen gas.
so, 0.1 mole of sulphuric acid gives 0.1 mole of hydrogen.
volume of hydrogen = 0.1 mol × 22.4 L/mol = 2.24 L
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Answer:
2AI°(S)+3H²SO⁴(aq)→AI²(SO⁴)³(aq)+3H²o(g)
2AI°(S)+3H²SO⁴(aq)→AI²(SO⁴)³(aq)+3H²o(g)6H+e-→6H°
2AI°(S)+3H²SO⁴(aq)→AI²(SO⁴)³(aq)+3H²o(g)6H+e-→6H°2AI°-6e-