Question

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)

Answer 1

Step-by-step explanation

Given that;

The mass of Al2(SO4)3 is 500 grams

The mass of Ca(OH)2 is 450 grams

The mass of CaSO4 is 596 grams

Follow the steps below to find the limiting reactant of the reaction

Step 1; Write the balanced equation of the reaction

`\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))`

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

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

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

`\begin{gathered} \text{ Mole = }\frac{500}{\text{ 342.15}} \\ \text{ Mole = 1.461 moles} \end{gathered}`

For Ca(OH)2

`\begin{gathered} \text{ Mole = }\frac{\text{ 450}}{\text{ 74.093}} \\ \text{ Mole = 6.073 moles} \end{gathered}`

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

`\begin{gathered} \text{ For Al}_2(SO_4)_3 \\ \text{ The mole ratio = }(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}`

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

`\text{ The limiting reactant of the reaction is Al}_2(SO_4)_3`

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