If 12.0 g of MgSO4 react with 15.0.g of BaCl2, what is the limiting reactant and what mass of BaSO4 is produced?

1 Answer
Apr 23, 2014

The limiting reactant is BaCl₂. The reaction produces 16.8 g of BaSO₄.

The limiting reagent of a reaction is the reactant gives the smallest amount of product. It is the only chemical that we use to calculate the theoretical yield of product.

Balance Equation:

The balanced equation is

MgSO₄ + BaCl₂ → MgCl₂ + BaSO₄

Identify Limiting Reactant

Molar masses are

MgSO₄ = 120.4 g/mol; BaCl₂ = 208.2 g/mol; BaSO₄ = 233.4 g/mol

Calculate Moles of BaSO₄ from Each Reactant

Moles of BaSO₄ from MgSO₄:

12.0 g MgSO₄ × #(1" mol MgSO₄")/(120.4" g MgSO₄") × (1" mol BaSO₄")/("1 mol MgSO₄")# = 0.0997mol BaSO₄

Moles of BaSO₄ from BaCl₂:

15.0 g BaCl₂ × #(1" mol BaCl₂")/(208.2" g BaCl₂") × (1" mol BaSO₄")/(1" mol BaCl₂")# = 0.0720 mol BaSO₄

BaCl₂ is the limiting reactant, because it gives the smaller amount of product.

Mass of BaSO₄:

0.0720 mol BaSO₄ × #(233.4" g BaSO₄")/(1" mol BaSO₄")# = 16.8 g BaSO₄