How many milliliters of 0.125 M FeCl3 are needed to react with an excess of Na2S to produce 3.75 g of Fe2S3 if the percent yield for the reaction is 75.0%?

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

54.1 mL

385 mL

216 mL

96.2 mL

185 mL

0.912 mL

Explanation:

3 Na2S(aq) + 2 FeCl3(aq) → Fe2S3(s) + 6 NaCl(aq)

FeCl3 is the limiting reactant.

Number of moles of iron III sulphide produced= 3.75g/87.92 g/mol = 0.043 moles

Hence actual yield of Iron III sulphide = 0.043 moles

Theoretical yield of Iron III sulphide = actual yield ×100%/ %yield

Theoretical yield of iron III sulphide= 0.043 ×100/75 = 0.057 moles of Iron III sulphide

From the reaction equation,

2moles of iron III chloride produced 1 mole of iron III sulphide

x moles of iron III chloride, will produce 0.057 of iron III sulphide

x= 2× 0.057= 0.114 moles of iron III chloride

But

Volume= number of moles/ concentration

Volume= 0.114/0.125

Volume= 0.912 mL