If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of 1 kg of it will be

Question

If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of 1 kg of it will be (A) 2.7 kg (B) 1.8 kg (C) 3.58 kg (D) 4.5 kg

Tardigrade Admin · Accepted Answer

Combustion of butane takes place as C 4 H 10+(13/2) O 2 longrightarrow 4 CO 2+5 H 2 O 4 × 12+10 × 1 =48+10 = 58 g Now, 58 kg C 4 H 10 require O 2 =(13/2) × 32=208 kg ∴ 1 kg C 4 H 10 require O 2 =(208/58)=3.58 kg

Q. If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of $1 k g$ of it will be

2.7 kg

1.8 kg

3.58 kg

4.5 kg

Combustion of butane takes place as
$C_{4} H_{10} + \frac{13}{2} O_{2} ⟶ 4 C O_{2} + 5 H_{2} O$
$4 \times 12 + 10 \times 1$
$= 48 + 10$
$= 58 g$
Now, $58 k g C_{4} H_{10}$ require $O_{2}$
$= \frac{13}{2} \times 32 = 208 k g$
$∴ 1 k g C_{4} H_{10}$ require $O_{2}$
$= \frac{208}{58} = 3.58 k g$

Q. If LPG cylinder contains mixture of butane and isobutane, then the amount of oxygen that would be required for combustion of 1kg of it will be