If 1g of Ba reacts with 1.8g Al (SO4) 3, what is the excess reactant

Al₂ (SO₄) ₃ is the excess reactant.

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.