MCQ
Product $(D)$ will be


- A

- ✓

- C

- D








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$Ca{C_2} + 2{H_2}O \to Ca{(OH)_2} + {C_2}{H_2}$
${C_2}{H_2} + {H_2} \to {C_2}{H_4}$
$n({C_2}{H_4}) \to {( - C{H_2} - C{H_2} - )_n}$
The amount of polyethylene obtained from $64.1\, kg$ $Ca{C_2}$ is......$kg$