2 moles of Sodium sulfate is reacted with 3 moles of calcium chloride.

Question

2 moles of Sodium sulfate is reacted with 3 moles of calcium chloride.  

1. Write a balanced chemical equation, including states of all reactants and products.
2. Write a net ionic equation.
3. Determine the limiting reactant.
4. Determine the mass of the precipitate that should be produced if the reaction has a 75% yield.

Answer 1

Sodium sulfate = Na₂SO₄

Calcium chloride = CaCl₂

1). Na₂SO₄(aq) + CaCl₂(aq) ==> 2NaCl(aq) + CaSO₄(s) ... balanced equation

2). Ca²⁺(aq) + SO₄^2-(aq) ==> CaSO₄(s)

3). Easiest way to find limiting reactant is to divide moles of each reactant by its coefficient in the equation:

2 moles Na₂SO₄ ÷ 1 = 2

3 moles CaCl₂ ÷ 1 = 3

Limiting reactant = Na₂SO₄

4). Because Na₂SO₄ is limiting, it alone will determine the amount of product. At 100% yield, you have...

2 mol Na₂SO₄ x 1 mol CaSO₄/ mol Na₂SO₄ x 136 g CaSO4/mol = 272 g CaSO₄

At 75% yield you'd have 0.75 x 272 g = 204 g CaSO₄