Aluminum hydroxide reacts with sulfuric acid as follows: 2Al(OH)3+3H2SO4âAl2(SO4)3+6H2O.Which reagent is the limiting reactant when 0.550 mol Al(OH)3 and 0.550 mol H2SO4 are allowed to react?How many moles of Al2(SO4)3 can form under these conditions?
How many moles of the excess reactant remain after the completion of the reaction?

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Xema8 Xema8 · Answer 1 · 2020-11-16T04:50:03+01:00

Answer:

Explanation:

2Al(OH)₃ + 3H₂SO₄ = Al₂(SO₄)₃ + 6H₂O

2 moles 3 moles

2 moles of Al(OH)₃ reacts with 3 moles of H₂SO₄

.55 moles of Al(OH)₃ reacts with .55 x 1.5 moles of H₂SO₄

moles of H₂SO₄ required = .825 moles

moles available = .55 moles , so H₂SO₄ is the limiting reagent .

3 moles of H₂SO₄ yields 1 moles of Al₂(SO₄)₃

.55 moles of H₂SO₄ yields 1 x .55 / 3 moles of Al₂(SO₄)₃

Al₂(SO₄)₃ produced = 1 x .55 / 3 moles = .1833 moles .

Al(OH)₃ reacted = 2 x .55 / 3 = .367 moles

excess Al(OH)₃ remaining = .5 - .367

= .133 moles .