We react 180.0g Fe2P with 140.0g of S according to the following reaction: 4FeP(s) + 18S(s)...

Question

Question

We react 180.0g Fe2P with 140.0g of S according to the following reaction:

4FeP(s) + 18S(s) -----> 8FeS(s) + P4S10(s)

What is the limiting reactant and how much FeS will be produced (in grams)?

How much of the excess reactant will we have left over in grams?

Answer 1

Answer #1

4Fe2P(s) + 18S(s) -----> 8FeS(s) + P4S10(s)

No of mol of Fe2P = 180/142.6638 = 1.26 mol

No of mol of S = 140/32 = 4.375 mol

limiting reactant = s

4 mol Fe2p = 18 mol S = 8 mol Fes

Mass of Fes formed = 4.375*(8/18)*87.9 = 170.91 grams

excess reactant = 180 - (4.375*(4/18)*142.7) = 41.264 grams

Answer 2

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