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How many milliliters of 0.200 M FeCl3 are needed to react with an excess of Na2S to produce 0.690 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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