189 g of sodium sulfate react with 147 g of silver nitrate, determine the mass of...
189 g of sodium sulfate react with 147 g of silver nitrate, determine the mass of the products
Homework Answers
mass of Na2SO4 = 189 g
mass of AgNO3 = 147 g
2 AgNO3 + Na2SO4 ---------------> Ag2SO4 + 2 NaNO3
339.75 g 142.04 g 311.8 170 g
147 g 189 g
339.75 g AgNO3 ---------------> 142. g Na2SO4
147 g AgNO3 ------------> ??
mass of Na2SO4 needed = 147 x 142 / 339.8 = 61.5 g
but we have 189 g so . this is excess .
limiting reagent is AgNO3 .
mass of products formed :
mass of Ag2SO4 formed = 147 x 311.8 / 339.75
= 135 g
mass of NaNO3 formed = 147 x 170 / 339.75
= 73.6 g
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