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How many millilitres of 0.200 mol L-1 FeCl3(aq) are needed to react with an excess of Na2S(aq) to produce 1.38 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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