Find out the amount of polythene formed from 20 kg of calcium carbide from the following reactions.
$C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2} H C$
$C H \equiv C H + H_{2} p d . B a S O_{4} ⟶ C H_{2} = C H_{2}$
$n C H_{2} = C H_{2} ⟶ {(- C H_{2} - C H_{2} -)}_{n} P o l y t h e n e$

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Solution

20.0kg of $C a C_{2}$ $= \frac{20.0 \times 10^{3}}{64} = 312.5$ moles of $C a C_{2}$
From the above reactions it is evident that 1 mol $C a C_{2}$ gives 1 mol of acetylene, of which n moles combines with n moles of H2 to give polyethylene (n molecules of $C_{2} H_{4}$ or polymerized $C_{2} H_{4}$ ).
Hence 312.5 moles of $C a C_{2}$ will give 312.5 moles of $C_{2} H_{4}$ $= 312.5 \times 28 = 8750 g$ or $8.75 k g$
OR the polyethylene formed is $8.75 k g$ .

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20

C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}

H C \equiv C H + H_{2} P d \cdot B a S O_{4} - ----- \to C H_{2} = C H_{2}

n C H_{2} = C H_{2} \to (- C H_{2} - C H_{2} -)_{n} P o l y t h e n e

C a C_{2} + 2 H_{2} O \to C a (O H)_{2} + C_{2} H_{2}

H C \equiv C H + H_{2} P d - B a S O_{4} \to C H_{2} = C H_{2}

n C H_{2} = C H_{2} \to (C H_{2} - C H_{2})_{n}

Calculate the amount of polythene formed from 20kg of Calcium Carbide from the reaction given below

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ (- C H_{2} - C H_{2} -)_{n}

C a C_{2}

C a C_{2} + 2 H_{2} O ⟶ C a {(O H)}_{2} + C_{2} H_{2}

C_{2} H_{2} + H_{2} ⟶ C_{2} H_{4}

n C_{2} H_{4} ⟶ {(- {C H}_{2} - {C H}_{2} -)}_{n}

64.1 k g

C a C_{2}

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