Formation of polyethylene from calcium carbide takes place as follows

$Ca{C}_{2} + 2{H}_{2}O \to Ca ( OH{)}_{2} + {C}_{2}{H}_{2}$

${C}_{2}{H}_{2} + {H}_{2} \to {C}_{2}{H}_{4}$

$n ( {C}_{2}{H}_{4} ) \to ( - C{H}_{2} - C{H}_{2} - {)}_{n}$

The mass of polyethylene obtained in kg from 64 kg $Ca{C}_{2}$ is

Question

Formation of polyethylene from calcium carbide takes place as follows

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 mass of polyethylene obtained in kg from 64 kg

Ca{C}_{2}

is

Quantrex Academy · Accepted Answer

Correct answer: 28

Ca C _ 2 _ 64g + 2 H _ 2 O → Ca ( OH ) _ 2 + C _ 2 H _ 2 C _ 2 H _ 2 + H _ 2 → C _ 2 H _ 4 _ 28g 64g of Ca C _ 2 gives 28g of ethylene ∴ 64kg of Ca C _ 2 will give 28kg of polyethylene

Formation of polyethylene from calcium carbide takes place as...

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