Answer to Question #229457 in Chemistry for Mike

Question

A quality control analyst mixes 30.0 ml of 0.255M BaCl2 with excess AgNO3 causing Angel to precipitate and subsequently isolated 2.01 g of AgCl. Show a balanced equation for the reaction and calculate the percent yield of the reaction.

Answer 1

BaCl₂ + 2AgNO₃ =2AgCl + Ba(NO₃)₂

According to the equation, n (AgNO₃) = n (AgCl) = 2 x n (BaCl₂)

C_M = n / V

n = C_M x V

n (BaCl₂) = 0.255 x 0.03 = 0.008 mol

n (AgCl) = 0.008 x 2 = 0.016 mol

n = m / M

M (AgCl) = 143.32 g/mol

m (AgCl) = n x M = 0.016 x 143.32 = 2.29 g

% = 2.01/2.29 x 100 = 87.8%