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How many mL of .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%? 3Na2S + 2FeCl3 = Fe2S3 + 6NaCl
a) 50.9 mL
b) 86.0 mL
c) 102 mL
d) 203 mL
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