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2al(oh)3 +3h2so4 = Al2(SO4)3 + 6h2o 0.450 mol Al(OH)3 and 0.550 mol H2SO4 How many moles of Al2(SO4)3 can form under these conditions? (c) how many moles of the excess reactant remain after the completion of the reaction?
Accepted Answer
The balanced chemical equation shows that 2 moles of Al(OH)3 react with 3 moles of H2SO4 to produce 1 mole of Al2(SO4)3.
To determine the limiting reactant, we can calculate the moles of Al2(SO4)3 that could be produced from each reactant:
From Al(OH)3: 0.450 mol Al(OH)3
(1 mol Al2(SO4)3 / 2 mol Al(OH)3) = 0.225 mol Al2(SO4)3
From H2SO4: 0.550 mol H2SO4
(1 mol Al2(SO4)3 / 3 mol H2SO4) = 0.183 mol Al2(SO4)3
Since H2SO4 produces less Al2(SO4)3, it is the limiting reactant. Therefore, 0.183 moles of Al2(SO4)3 can form.
To calculate the excess reactant, we first need to determine how much of the excess reactant was consumed:
H2SO4 consumed: 0.183 mol Al2(SO4)3
(3 mol H2SO4 / 1 mol Al2(SO4)3) = 0.549 mol H2SO4
Excess H2SO4: 0.550 mol H2SO4 (initial) - 0.549 mol H2SO4 (consumed) = 0.001 mol H2SO4
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