Al(OH)
William W. answered • 05/26/23
Math and science made easy - learn from a retired engineer
Starting with the unbalanced reaction:
Al2(SO4)3 + Ca(OH)2 → Al(OH)3+ CaSO4
Step 1) - Balance the equation:
Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3+ 3CaSO4
Step 2) - Calculate the moles of Al2(SO4)3 (aluminum sulfate) and Ca(OH)2 (calcium hydroxide)
First find the molar mass of each:
Al: 26.982 x 2 = 53.964
S: 32.066 x 3 = 96.198
O: 15.999 x 12 = 191.988
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Al2(SO4)3 = 342.15 g/mol
Ca: 40.078 x 1 = 40.078
O: 15.999 x 2 = 31.998
H: 1.008 x 2 = 2.016
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Ca(OH)2 = 74.092 g/mol
Calculate moles of Al2(SO4)3 : (13.8 g)/(342.15 g/mol) = 0.04033 moles
Calculate moles of Ca(OH)2 : (2.04 g)/(74.092 g/mol) = 0.02753 moles
Step 3) - Find the limiting reactant:
According to the balanced chemical reaction equation, it takes 1 mole of Al2(SO4)3 to mix with 3 moles of Ca(OH)2 to make the chemical reaction occur. In this case you have extra Al2(SO4)3 since the 0.02753 moles of Ca(OH)2 would only require 0.02753/3 or 0.00918 moles of Al2(SO4)3 and you have 0.04+ moles of it. So Ca(OH)2 is the limiting reactant.
