What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

Question

What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene? (A) 2.8 kg (B) 6.4 kg (C) 9.6 kg (D) 96 kg

Tardigrade Admin · Accepted Answer

underset(12 ×2 +4)C 2 H 4+ underset96 kg3 O 2 arrow 2 CO 2+2 H 2 =28 kg ∵ Weight of oxygen required for the Complete combustion of 28 kg ethylene =96 kg ∵ Weight of oxygen required for the combustion of 2.8 kg ethylene. =(96 × 2.8/28)=9.6 kg

Q. What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

2.8 kg

6.4 kg

9.6 kg

96 kg

$(12 \times 2 + 4) C_{2} H_{4} + 96 k g 3 O_{2} \to 2 C O_{2} + 2 H_{2}$
$= 28 k g$
$∵$ Weight of oxygen required for the Complete combustion of $28 k g$ ethylene $= 96 k g$
$∵$ Weight of oxygen required for the combustion of $2.8 k g$ ethylene.
$= \frac{96 \times 2.8}{28} = 9.6 k g$

Q. What is the weight of oxygen required for the complete combustion of 2.8 kg of ethylene?

2.8 kg

6.4 kg

9.6 kg

96 kg

(12×2+4)C2​H4​​+96kg3O2​​→2CO2​+2H2​ =28kg ∵ Weight of oxygen required for the Complete combustion of 28kg ethylene =96kg ∵ Weight of oxygen required for the combustion of 2.8kg ethylene. =2896×2.8​=9.6kg

$(12 \times 2 + 4) C_{2} H_{4} + 96 k g 3 O_{2} \to 2 C O_{2} + 2 H_{2}$
$= 28 k g$
$∵$ Weight of oxygen required for the Complete combustion of $28 k g$ ethylene $= 96 k g$
$∵$ Weight of oxygen required for the combustion of $2.8 k g$ ethylene.
$= \frac{96 \times 2.8}{28} = 9.6 k g$