azireyathurmond1
28.11.2019 •
Chemistry
0.450 mol of aluminum hydroxide is allowed to react with 0.550 mol of sulfuric acid; the reaction which ensues is: 2al(oh)3(s) + 3h2so4(aq) > al2(so4)3(aq) + 6h2o(l) how many moles of h2o can form under these conditions?
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Ответ:
The answer to your question is 1.1 moles of water
Explanation:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
0.45 mol 0.55 mol ?
Process
1.- Calculate the limiting reactant
Theoretical proportion
Al(OH)₃ / H₂SO₄ = 2/3 = 0.667
Experimental proportion
Al(OH)₃ / H₂SO₄ = 0.45 / 0.55 = 0.81
From the proportions, we conclude that the limiting reactant is H₂SO₄
2.- Calculate the moles of H₂O
3 moles of H₂SO₄ 6 moles of water
0.55 moles of H₂SO₄ x
x = (0.55 x 6) / 3
x = 3.3 / 3
x = 1.1 moles of water
Ответ:
Kinetic energy
Explanation: