How many milliliters of 0.200 M FeCl₃ are needed to react with an excess of Na₂S to produce 2.75 g of Fe₂S₃ if the percent yield for the reaction is 65.0%?
3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)

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The Answer of How many milliliters of 0.200 M FeCl₃ are needed to react with an excess of Na₂S to produce 2.75 g of Fe₂S₃ if the percent yield for the reaction is 65.0%? 3 Na₂S(aq)+ 2 FeCl₃(aq)→ Fe₂S₃(s)+ 6 NaCl(aq)

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