A 500 g sample of Al2(SO4)3 is reacted with 450 g of Ca(OH)2. A total of 596 g of CasO4 isproduced. What is the limiting reactant in this reaction, and how many moles of excess reactantare unreacted? Al2(SO4)3 (ag) + 3Ca(OH)2 (ag) -> 2Al(OH)3 (s) + 3CaSO4 (s)

[tex]\text{ Al}_2(SO_4)_{3(aq)}+\text{ 3Ca\lparen OH\rparen}_{2(aq)}\text{ }\rightarrow\text{ 2Al\lparen OH\rparen}_{3(s)}\text{ + 3CaSO}_{4(s)}[/tex]

In the reaction above, 1 mole Al2(SO4)3 reacts with 3 moles Ca(OH)2 to give 2 moles Al(OH)3 and 3 moles CaSO4

Step 2; Determine the number of moles using the below formula

[tex]\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}[/tex]

Recall, that the molar mass of Al2(SO4)3 is 342.15 g/mol and the molar mass of Ca(OH)2 is 74.093 g/mol

For Al2(SO4)3

[tex]\begin{gathered} \text{ Mole = }\frac{500}{\text{ 342.15}} \\ \text{ Mole = 1.461 moles} \end{gathered}[/tex]

For Ca(OH)2

[tex]\begin{gathered} \text{ Mole = }\frac{\text{ 450}}{\text{ 74.093}} \\ \text{ Mole = 6.073 moles} \end{gathered}[/tex]

Step 3; Find the limiting reactant of the reaction

To find the limiting reactant of the reaction, divide the moles of the reactant by the co-efficient of the compound

[tex]\begin{gathered} \text{ For Al}_2(SO_4)_3 \\ \text{ The mole ratio = }\frac{1.461}{1} \\ \text{ The mole ratio = 1.461 mol/wt} \\ \\ \text{ For Ca\lparen OH\rparen}_2 \\ \text{ The mole ratio = }\frac{\text{ 6.073}}{3} \\ \text{ The mole ratio = 2.024 mol/wt} \end{gathered}[/tex]

Since the limiting reactant of the reaction is the reactant with the lowest number of mol/wt, then Al2(SO4)3 is the limiting reactant

[tex]\text{ The limiting reactant of the reaction is Al}_2(SO_4)_3[/tex]

The excess reactant of the reaction is Ca(OH)2

Therefore, the no of moles of the excess reactant that is unreacted is

6.073 - 1.461 = 4.612 moles

Hence, the number of moles of the excess reactant that is unreacted is 4.621 moles