Question

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

Answer 1

Answer:

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.