Homework Answers
|
Starting material/product |
Molecular weight |
Amount taken |
moles |
Equivalent |
Product obtained |
|
p-Anisaldehyde |
136.15 (d=1.119) |
1 mL (1.119 gm) |
0.00821 mol |
1 |
|
|
Acetophenone |
120.15 (d=1.028) |
1 mL (1.028 gm) |
0.00855 mol |
1.04 |
|
|
Product |
239.30 |
1.232 gm |
|||
p-anisaldehyde is the limiting reagent in the reaction
136.15 gm of p-anisaldehyde gives 239.30 gm of Product
Therefore 1.119 gm of p-anisaldehyde will give x gm of product
Theoretical yield of the reaction = (1.119*239.30)/136.50 = 1.96 gm
Percent yield of the reaction = (1.232/1.96)*100 = 62.85%
Add Answer to:
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