Asked by Melody
how many millimeters of 0.200M FeCL3 are needed to react with an excess of Na2S to produce 1.38g of Fe2S3 if the percent yield for the reaction is 65.0%
Answers
Answered by
DrBob222
2FeCl3 + 3Na2S ==> Fe2S3 + 6NaCl
mols Fe2S3 = grams/molar mass = ? That's the mols you want if the reaction were 100% but it is only 65%; therefore, you want ? mol/0.65 = mols needed at 65% yield.
Using the coefficients in the balanced equation, convert mols Fe2S3 to mols
FeCl3.
Then M FeCl3 = mols FeCl3/L FeCl3. You know M and mols, solve for L and convert to mL.
mols Fe2S3 = grams/molar mass = ? That's the mols you want if the reaction were 100% but it is only 65%; therefore, you want ? mol/0.65 = mols needed at 65% yield.
Using the coefficients in the balanced equation, convert mols Fe2S3 to mols
FeCl3.
Then M FeCl3 = mols FeCl3/L FeCl3. You know M and mols, solve for L and convert to mL.
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