% yield = 80.719
Given
22.0 g of Mgl₂
25.0 g of Mg
25.0 g of l₂
Required
The percent yield
Solution
Reaction
Mg + I₂⇒ MgI₂
mol Mg = 25 g : 24.305 g/mol = 1.029
mol I₂ = 25 g : 253.809 g/mol = 0.098
Limiting reactant = I₂
Excess reactant = Mg
mol MgI₂ based on I₂, so mol MgI₂ = 0.098
Mass MgI₂ (theoretical):
= mol x MW
= 0.098 x 278.114
= 27.255 g
% yield = (actual/theoretical) x 100%
% yield = (22 / 27.255) x 100%
% yield = 80.719
