Answer to Question #56144 in General Chemistry for Michael Scheller

Question

Strontium chloride reacts with silver nitrate according to the equation SrCl2 + 2AgNO3 ➡️ 2AgCl + Sr(NO3)2 What is the percentage yield for a reaction in which 2.50 g of AgCl is obtained from 1.50 g of strontium chloride and a slight excess of silver nitrate?

Answer 1

n(SrCl₂) = m(₂)/MW(₂) = 1.50 g/158.53 g mo^l-1 = 9.46 mmol
m(AgCl _th) = 2 n(₂) MW(AgCl) = 2*9.46 mmol * 143.32 g mol^-1 = 2.71 g
w(AgCl) = m(AgCl)/m(AgCl _th) = 2.50 g/2.71 g = 92.2 %