Stoichiometry in Solutions

Question

4.67g of solid silver nitrate are mixed into a 200.0mL solution of 0.500mol/L aqueous potassium sulfate. 



1. Calculate the moles of each reactant 

2. Identify which reactant is the limiting reactant

3. Calculate the moles of each product 

4. Calculate the theoretical yield of solid precipitate 

5. Assuming you get 3.88g of solid precipitate, calculate the percent yield 

Answer 1

You should always first write the correctly balanced equation:

2AgNO₃(aq) + K₂SO₄(aq) ==> 2KNO₃(aq) + Ag₂SO₄(s) ... balanced equation

1). moles of each reactant:

molar mass AgNO₃ = 170 g/mol

4.67 g AgNO₃ x 1 mol AgNO₃ / 170 g = 0.0275 mols AgNO₃

200.0 ml K₂SO₄ x 1 L / 1000 ml x 0.500 mol / L = 0.100 mols K₂SO₄

2). Limiting reactant. Divide moles by corresponding coefficient in balanced equation:

AgNO₃: 0.0275 mols / 2 = 0.014

K₂SO₄: 0.100 mols / 1 = 0.100

0.014 is less than 0.100 so AgNO₃ is LIMITING

3). moles of each product formed:

0.0275 mols AgNO₃ x 2 mols KNO₃ / 2 mol AgNO₃ = 0.0275 mols KNO₃ formed

0.0275 mols AgNO₃ x 2 mol Ag₂SO₄ / 2 mol AgNO₃ = 0.0138 mols Ag₂SO₄ formed

4). Theoretical yield of Ag₂SO₄:

molar mass Ag₂SO₄ = 312 g/mol

0.0138 mols Ag₂SO₄ x 312 g / mol = 4.29 g Ag₂SO₄

5). Percent yield:

%yield = actual yield / theoretical yield (x100%)

%yield = 3.88g / 4.29 g (x100%) = 90.4% yield