A) 14 g
B) 28 g
C) 64.5 g
D) 56 g
E) 7g
Correct Answer: A
Solution :
\[{{C}_{2}}{{H}_{5}}Cl\xrightarrow[{}]{dehydroha\log enation}{{C}_{2}}{{H}_{4}}\] \[(24+5+35.5)\] \[(24+4)\] g of\[{{C}_{2}}{{H}_{5}}Cl\]forms\[=28\text{ }g\text{ }{{C}_{2}}{{H}_{4}}\] \[\therefore \]\[32.25\text{ }g\text{ }of\text{ }{{C}_{2}}{{H}_{5}}Cl\] will form\[=\frac{28}{64.5}\times 32.25\] \[=14\,g\,{{C}_{2}}{{H}_{4}}.\]
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