Polyethylene is obtained from calcium carbide.

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

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

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

Therefore, the amount of polyethylene (in kg) obtained for 64 kg $Ca{C}_{2}$ is

Question

Polyethylene is obtained from calcium carbide.

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

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

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

Therefore, the amount of polyethylene (in kg) obtained for 64 kg

Ca{C}_{2}

is

Quantrex Academy · Accepted Answer

Correct answer: 28

Moles of Ca C _ 2 = 64 × 1 0 ^ 3 64 ≅ 1 × 1 0 ^ 3 So from the balanced chemical equation, moles of C _ 2 H _ 2 = moles of C _ 2 H _ 4 = moles of Ca C _ 2 = 1 × 1 0 ^ 3 So moles of polythene = (1)/(n) × 1 × 1 0 ^ 3 So weight of polythene = (1)/(n) × 1 × 1 × 28n kg = 28 kg

Polyethylene is obtained from calcium carbide. Ca C _ 2 + 2 H _ 2 O...

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