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

$2AgNO_{3(aq)}+K_2SO_{4(aw)}\to\>Ag_2SO_{4(s)}+2KNO_{3(aq)}$

Molar mass of AgNO$_3=169.87$

$K_2SO_4=174.259$

$Ag_2SO_4=311.799$

Part 1

Moles of AgNO$_3=\frac{4.67}{169.87}=0.02749\>moles$

Moles of $K_2SO_4=\frac{200}{1000}×0.5=0.1\>moles$

Part 2

$0.1×2=0.2>0.02749$

$\therefore$ $AgNO_3$ is the limiting reagent

Part 3

Moles of $Ag_2SO_4=\frac{1}{2}×0.02749$

$=0.01375$

Moles of $KNO_3=0.02749$

Part 4

Theoretical yield of $Ag_2SO_4=0.01375×311.799$

$=4.2859g$

Part 5

Percentage yield $=\frac{3.88}{4.2859}×100\%=90.53\%$