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.2M x V FeCl3=moles FeCl3 x 1mole Fe2S3/2mole FeCl3=moles of Fe2S3 x mm of Fe2S3/1 mole Fe2S3= g Fe2S3 x .65% yield.

2.75g Fe2S3/ .65= 4.23g Fe2S3/ 207.91= .02035 x 2mole FeCl3=.0407 moles FeCl3/ .2M FeCl3= .2035 L x 1000= 203.5 ml

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Q: 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 percent?
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