Answer to Question #115716 in General Chemistry for sam

Question

40g of barium hydroxide is reacted with 40g of ammonium sulfate, what is the limiting reactant

Answer 1

Solution:

The balanced chemical equation:

Ba(OH)₂ + (NH₄)₂SO₄ = BaSO₄ + 2NH₃ + 2H₂O

The molar mass of Ba(OH)₂ is 171.34 g/mol

The molar mass of (NH₄)₂SO₄ is 132.14 g/mol

The molar mass of BaSO₄ is 233.38 g/mol

Solution path #1:

Use stoichiometry to calculate how much product is produced by each reactant (For example, BaSO₄).

The reactant that produces the lesser amount of product is the limiting reactant: in this case is barium hydroxide - Ba(OH)₂.

Solution path #2:

Calculate moles:

1) Moles of substance = Mass of substance / Molar mass of substance

2) Divide by coefficients of balanced equation:

barium hydroxide ⇒ 0.233 mol / 1 mol = 0.233

ammonium sulfate ⇒ 0.303 mol / 1 mol = 0.303

Barium hydroxide is the lower value. It is the limiting reagent.

Ba(OH)₂ is the limiting reactant.

Answer: Ba(OH)₂ is the limiting reactant.