Answer:
Al₂(SO₄)₃ is the excess reactant.
Step-by-step explanation:
Barium (Ba) react with Aluminium sulphate [Al₂(SO₄)₃] according to the following balanced equation:
- 3 Ba + Al₂(SO₄)₃ → 3 BaSO₄ + 2 Al
It is clear that 3 mol of Ba react with 1 mol of Al₂(SO₄)₃ to give 3 moles of BaSO₄
The limiting reactant is the reactant that produces the least amount of BaSO ₄.
The molar masses of each substance involved.
Ba : 137.3 g/mol
Al₂(SO₄)₃: 342.1 g/mol
BaSO ₄: 233.3 g/mol
Then we calculate no of moles of each reactant from the given mass.
As following:
no. of moles of Ba = (mass /molar mass) = (1 g / 137.3 g/mol) = 0.0073 mol
no. of moles of Al₂(SO₄)₃ = (mass /molar mass)
= (1.8 g / 342.1 g/mol) = 0.0053 mol
Then we calculate mol of product produced from each reactant
For BaSO₄
- 3 mol of BaSO₄ produced from → 3 mol of Ba
- ?? mol of BaSO₄ produced from → 0.0073 mol of Ba
no of moles of BaSO₄ from Ba = (0.0073 * 3) / 3 = 0.0073 mol
then converting moles of BaSO₄ into mass
mass of BaSO₄ = no of moles * molar mass = 0.0073 * 233.3 = 1.7 g
For Al₂(SO₄)₃
- 3 mol of BaSO₄ produced from → 1 mol of Al₂(SO₄)₃
- ?? mol of BaSO₄ produced from → 0.0053 mol of Al₂(SO₄)₃
no of moles of BaSO₄ from Al₂(SO₄)₃ = (0.0053 * 3) / 1 = 0.0159 mol
then converting moles of BaSO₄ into mass
mass of BaSO₄ = no of moles * molar mass = 0.0159 * 233.3 = 3.7 g
∴1 g of Ba produces the least amount of barium sulfate, so it is the limiting reactant and Al₂(SO₄)₃ is the excess reactant.