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Answer to Question #56144 in General Chemistry for Michael Scheller

Question #56144
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?
Expert's answer
n(SrCl2) = m(2)/MW(2) = 1.50 g/158.53 g mol-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 %

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