Find the limiting reactant, which is the reactant that forms the less amount of product.
Balanced equation: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
Mole to mole comparison:
Since moles of Al(OH)3 and H2SO4 are equal → limiting reactant = H2SO4
The moles of Al2(SO4)3 formed under the given conditions is:
0.183 mol Al2(SO4)3
You may want to reference (Pages 106 - 109)Section 3.7 while completing this problem.
Aluminum hydroxide reacts with sulfuric acid as follows: 2 Al(OH)3(s) + 3 H2SO4(aq) → Al2(SO4)3(aq) + 6 H2O(l)
0.550 moles Al(OH)3 and 0.550 moles H2SO4 are allowed to react. How many moles of the excess reactant remain after the completion of the reaction?
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