Answer to Question #148915 in Chemistry for Emily Heilbron

Question

In a reaction between 1.25 g of KCl and 2.00 g of AgNO3 a mass of 2.37 g of AgCl is recovered. What is the percent yield for this reaction?

Answer 1

1 KCl + 1 AgNO₃ = 1 AgCl + 1 KNO₃

n = m/M

M (KCl) = 74.6 g/mol

M (AgNO₃) = 169.9 g/mol

M (AgCl) = 143.3 g/mol

n (KCl) = 1.25/74.6 = 0.017 mol

n (AgNO₃) = 2.0 / 169.9 = 0.012 mol

AgNO₃ is the limiting reactant. Anticipated amount of AgCl is:

m (AgCl) = 0.012 x 143.3 = 1.69 g

% yield AgCl = 2.37 / 1.69 x 100 = 140% (which is impossible, check the incoming data).