How many millilitres of 0.200 mol L^-1 FeCl₃ are needed to react with an excess of Na₂S to produce 1.38 g of Fe₂S₃ if the percent yield for the reaction is 65.0%?
3Na₂S(aq) + 2FeCl₃(aq) → Fe₂S₃(s) + 6NaCl(aq)

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The Answer of How many millilitres of 0.200 mol L^-1 FeCl₃ are needed to react with an excess of Na₂S to produce 1.38 g of Fe₂S₃ if the percent yield for the reaction is 65.0%? 3Na₂S(aq) + 2FeCl₃(aq) → Fe₂S₃(s) + 6NaCl(aq)

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