If a 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 a 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) 2.50 × 103 g (B) 4.50 × 103 g (C) 1.80 × 103 g (D) 3.58 × 103 g

Tardigrade Admin · Accepted Answer

The standard enthalpy of combustion of butane

C_{4} H_{10}

(both

n

-and iso butanes are main constituents of cooking gas). representing the combustion of 1 mole of butane in the presence of oxygen, may be representing represented as follows

1 mol C_{4} H_{10} (g) + \frac{13}{12} m o l \frac{13}{2} O_{2} (g) \to 4 C O_{2} (g) + 5 H_{2} O (l)

;

Δ_{c} H^{\circ} = - 2658.0 k J m o l^{- 1}

(58 g) (\frac{13}{2} \times 32 = 208 g)

∴ O_{2}

required for

1 k g

butane

= \frac{208}{58} \times 1000 g

= 3.58 \times 1 0^{3} g

Q. If a $L PG$ 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

$2.50 \times 1 0^{3} g$

$4.50 \times 1 0^{3} g$

$1.80 \times 1 0^{3} g$

$3.58 \times 1 0^{3} g$

Q. If a 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

2.50×103g

4.50×103g

1.80×103g

3.58×103g

Q. If a $L PG$ 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

$2.50 \times 1 0^{3} g$

$4.50 \times 1 0^{3} g$

$1.80 \times 1 0^{3} g$

$3.58 \times 1 0^{3} g$