can someone help me determine the % yield. After 5mL of furan and 0.056g of N-ph

Question

can someone help me determine the % yield. After 5mL of furan and 0.056g of N-phenylmaleimide were placed in the vial, a solid formed and the mass of that solid was 0.051g. I don't know how to calculate the actual yield. Thank you.

0 N-phenylmaleimide Furan m/z-241.25 g/mol m/z-68.07 g/mol m/z - 173.17 g/mol 0.5 mL 56 mg density-0.936 g/mL So, weight of furan 0.5 x 0.936 0.468 grams No. of moles of N-pheynlmaleimide -56 mg/ 173.17-0.3234 mmoles No. of moles of Furan 468 mg/68.07- 6.97 mmoles Here, Both will react in 1:1 equivalents. No. of moles of N-phenylmaleimide present in limited quantity. So, the limiting reagent is N-phenylmaleimide No. of moles of product formed-0.3234 mmoles So, weight (theoretical) of product 0.3234 x 241.25 -78 mg

Explanation / Answer

here theoretical yield = 0.078 g (as calculated above)

as given in question that the mass of that solid formed was 0.051g.

So, actual yield = 0.051g.

now, you can percentage yield = (0.051/0.078)x100 = 65.38%

if you satisfied with the solution please like it.. thanks..

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