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How many milliliters of 0.200 M FeBr3 are needed to react with Li2S to produce 2.75 g of Fe2S3 if the percent yield for the reactionis65.0%?
3 Li2S(aq) + 2 FeBr3(aq) -> Fe2S3(s) +6 LiBr(aq)
2 Answers
- az_lenderLv 75 days ago
Molar mass of Fe2S3 is 87.91g/mol.
So the 2.75 grams in the problem would be 31.28 millimoles.
Therefore you need (31.28 millimoles)/0.65 of FeBr3,
which is to say, you need 48.12 millimoles of FeBr3 to be available.
From an 0.200 M solution, you need 5 x 48.12 mL = 242.4 mL.
Make it 243 mL if you want "at least" 2.75 grams of Fe2S3.
- hcbiochemLv 75 days ago
2.75 g Fe2S3 / 87.91 g/mol X (2 mol FeBr3 / 1 mol Fe2S3) X (1 L/0.200 mol FeBre) X (1000 mL/1 L) = 313 mL FeBr3
So, at 100% efficiency, you would require 313 mL. Because of the reduced % yield,
313 mL X 1.35 = 422 mL