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 is produced. What is the limiting reagent in this reaction, and how many moles of excess reagent are unreacted?

Answer 1

Aluminum sulfate is the limiting reagent, and 1.69 moles of calcium hydroxide remain unreacted as the excess reagent in the reaction.

Step-by-step explanation:

To determine the limiting reagent and excess reagent for the reaction between Al2(SO4)3 and Ca(OH)2, leading to the production of CaSO4, we first need to write the balanced chemical equation for this reaction:

Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3

Then, we convert the masses of the reactants to moles:

Molar mass of Al2(SO4)3 = 342.15 g/mol

Molar mass of Ca(OH)2 = 74.09 g/mol

Moles of Al2(SO4)3 = 500 g / 342.15 g/mol = 1.46 mol

Moles of Ca(OH)2 = 450 g / 74.09 g/mol = 6.07 mol

Based on stoichiometry, we need 3 moles of Ca(OH)2 for every mole of Al2(SO4)3 to react completely. Thus we calculate how much Ca(OH)2 is needed for 1.46 moles of Al2(SO4)3:

3 moles Ca(OH)2 / 1 mole Al2(SO4)3 × 1.46 moles Al2(SO4)3 = 4.38 moles Ca(OH)2 required

Since we have 6.07 moles of Ca(OH)2, which is more than the 4.38 moles required, the excess reagent is Ca(OH)2 and Al2(SO4)3 is the limiting reagent. The amount of excess Ca(OH)2 is:

6.07 moles - 4.38 moles = 1.69 moles Ca(OH)2 excess

Answer 2

The chemical reaction is as follows:
Al2(SO4)3 + 3Ca(OH)2 → 3CaSO4 + 2Al(OH)3
Step-by-step explanation:
1mol of Al2(SO4)3 will react with 3mol Ca(OH)2 to produce 3mol CaSO4 and 2 mol Al(OH)3.
First we have to find the number of moles of Al2(SO4)3 :
Number of moles = Mass/ Molar mass
Mass of Al2(SO4)3 = 500g
Molar mass of Al2(SO4)3 = 342.15 g/mol
Number of moles = 500/342.15
Number of moles = 1.461 mol Al2(SO4)3
Multiplying the coeffecient of Ca(OH)2 with 1.461:
= 3*1.461 = 4.383 mol Ca(OH)2
Now we have to find the number of moles of Ca(OH)2:
Mass of Ca(OH)2 = 450g
Molar mass of Ca(OH)2 = 74.09 g/mol
Number of moles = 450/74.09
Number of moles = 6.074 mol Ca(OH)2
We need 4.383mol to react completely with the Al2(SO4)3, so the Ca(OH)2 is in excess, and the Al2(SO4)3 is the limiting reactant.
Excess unreacted: 6.074-4.383 = 1.69mol Ca(OH)2 unreacted .