Using the balanced equation above, determine the amount of excess reactant remaining when 99.7 g of Na2SO4 reacts with 75.0 g of AgNO3.

Question

Question

Na2SO4(aq) + 2AgNO3(aq) → Ag2SO4(s) + 2NaNO3(aq)

Answer 1

Answer #1

Na2SO4(aq) + 2AgNO3(aq) --------> Ag2SO4(s) + 2NaNO3(aq)

molecular weight of Na2SO4 = 142 g/mol

molecular weight of AgNO3 = 169.87 g/mol

99.7 g of Na2SO4 = 99.7/142 = 0.70 mol

75.0 g of AgNO3 = 75/169.87 = 0.44 mol

according to the reaction 1 mol of Na2SO4 reacts with 2 mol of AgNO3

so 0.22 mol of Na2SO4 reacts with 0.44 mol of AgNO3

so 0.48 mol of Na2SO4 will remain unreacted

amount of excess reactant = 0.48*142 = 68.16 g

Answer 2

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