Asked by Emmanuel
How many grams of commercial C2H4O2 (97% pure by mass) must be allowed to react with an excess of PCl3 to produce 75.0g of C2H3OCL if the reaction has a 78.2% yield?
Answers
Answered by
DrBob222
C2H4O2 = A with molar mass approx 60.
C2H3OCl = B with molar mass approx 78,
Note that all of these numbers are estimates so you should recalculate throughout for more accurate values.
A + PCl3 = B
Since the yield is 78.2% and not 100%, you will need to prepare 75.0/0.782= approx 95 g in order to obtain 75 g.
mols B = 95/78 = about 1.2
1 mol B is obtained from 1 mol A; therefore, you must start with 1.2 mols A.
Convert mols A to grams A by 1.2 mols A x 60 = approx 73 g but that's true only for 100% A and it is only 97% pure so you must start with 73/0.97 = approx ? g.
C2H3OCl = B with molar mass approx 78,
Note that all of these numbers are estimates so you should recalculate throughout for more accurate values.
A + PCl3 = B
Since the yield is 78.2% and not 100%, you will need to prepare 75.0/0.782= approx 95 g in order to obtain 75 g.
mols B = 95/78 = about 1.2
1 mol B is obtained from 1 mol A; therefore, you must start with 1.2 mols A.
Convert mols A to grams A by 1.2 mols A x 60 = approx 73 g but that's true only for 100% A and it is only 97% pure so you must start with 73/0.97 = approx ? g.
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