Polyethylene can be produced from CaC2 according to the following sequence of reactions-CaC2 - H2 O - CaO - C2 H2 Yield - 70 C2 H2 - H2 - C2 H4 Yield - 50 Calculate the mass of polyethylene which can be produced from 20 kg of CaC2-

CaC_2 + H_2 O → CaO + C_2 H_2 (i) C_2 H_2 + H_2 → C_2 H_4 (ii) nC_2 H_4 →(CH_2 CH_2 )_n (iii) In reaction (i), 64 g of CaC_2 produces 26 g of C_2 H_2 20 k ...

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Standard XII

Chemistry

Stoichiometric Calculations

Polyethylene ...

Question

Polyethylene can be produced from $C a C_{2}$ according to the following sequence of reactions.
$C a C_{2} + H_{2} O \to C a O + C_{2} H_{2} (Yield = 70 %)$
$C_{2} H_{2} + H_{2} \to C_{2} H_{4} (Yield = 50 %)$ $n C_{2} H_{4} \to (C H_{2} C H_{2})_{n} (Yield = 20 %)$
Calculate the mass of polyethylene which can be produced from $20 kg$ of $C a C_{2}$ .

612 kg

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558 g

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612 g

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558 kg

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Solution

The correct option is C $612 g$
$C a C_{2} + H_{2} O \to C a O + C_{2} H_{2} (i)$
$C_{2} H_{2} + H_{2} \to C_{2} H_{4} (ii)$
$n C_{2} H_{4} \to (C H_{2} C H_{2})_{n} (iii)$
In reaction (i),
$64 g$ of $C a C_{2}$ produces $26 g$ of $C_{2} H_{2}$
$20 kg$ will produce $= \frac{26}{64} \times 20 \times 10^{3} = 8.125 \times 10^{3} g$
Yield of the reaction is given as 70 %
Actual amount formed $= 8.125 \times 10^{3} \times 0.7 = 5.687 \times 10^{3} g$
In reaction (ii),
$26 g$ of $C_{2} H_{2}$ produces $28 g$ of $C_{2} H_{4}$
$5.687 \times 10^{3} g$ will produce = $\frac{28}{26} \times 5.687 \times 10^{3} = 6.125 \times 10^{3} g$
Yield of the reaction is given as 50 %
Actual amount of $C_{2} H_{4}$ formed $= 6.125 \times 10^{3} \times 0.50 = 3.06 \times 10^{3} g$
In reaction (iii),
$28 \times n g$ of $C_{2} H_{4}$ forms $28 \times n g$ of polyethylene
$3.06 \times 10^{3} g$ will form $= 3.06 \times 10^{3} g$ of polyethylene
Yield of the reaction is given as 20 %
Actual yield $= 3.06 \times 10^{3} \times 0.20 = 612 g$

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Similar questions

Q. Polyethylene can be produced from

C a C_{2}

according to the following sequence of reactions.

C a C_{2} + H_{2} O \to C a O + C_{2} H_{2} (Yield = 70 %)

C_{2} H_{2} + H_{2} \to C_{2} H_{4} (Yield = 50 %)

n C_{2} H_{4} \to (C H_{2} C H_{2})_{n} (Yield = 20 %)

Calculate the mass of polyethylene which can be produced from

20 kg

C a C_{2}

C a C_{2} + H_{2} O \to C a O + H C \equiv C H

n (C H \equiv C H) + n H_{2} \to - (C H_{2} - C H_{2})_{n} -

Polyethylene can be produced from calcium carbide according to the given sequence of reactions.
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C a C_{2} + 2 H_{2} O \to C a (O H)_{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 64.0 kg of

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Q. The formation of polyethylene from calcium carbide takes places as follows :

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

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C a C_{2}

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Q. Formation of polyethene from calcium carbide takes place as follows:

C a C_{2} + H_{2} O \to C a (O H)_{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 possibly obtainable from 64.0 kg

C a C_{2}

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Polyethylene can be produced from CaC2 according to the following sequence of reactions. CaC2+H2O→CaO+C2H2 (Yield = 70 %) C2H2+H2→C2H4 (Yield = 50 %) nC2H4→(CH2CH2)n (Yield = 20 %) Calculate the mass of polyethylene which can be produced from 20 kg of CaC2.