A gas mixture of 3 litres of propane C 3 H 8 and butane C 4 H 10 on complete combustion at 250 produced 10 litre CO 2- Find out the composition of gas mixture Propane - ButaneA- 3- 1B- 1- 2 - 1- 1D- 2- 1

Balanced equation for combustion reactions: aC_3H_8 + 5O_2 →3a3CO_2 + 4H_2O (3 a)C_4H_10 + 132 O_2 →4(3 a)4CO_2 + 5H_2O nbsp;Applying Gay Lussac's law, nb ...

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Standard XII

Chemistry

Avogadro's Law & Gay Lussac's Law

A gas mixture...

Question

A gas mixture of 3 litres of propane $(C_{3} H_{8})$ and butane $(C_{4} H_{10})$ on complete combustion at $25^{o}$ produced 10 litre $C O_{2} .$ Find out the composition of gas mixture (Propane : Butane)

3 : 1

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1 : 2

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2 : 1

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1 : 1

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Solution

The correct option is C $2 : 1$

Balanced equation for combustion reactions:
$C_{3} H_{8} a + 5 O_{2} \to 3 C O_{2} 3 a + 4 H_{2} O$
$C_{4} H_{10} (3 - a) + \frac{13}{2} O_{2} \to 4 C O_{2} 4 (3 - a) + 5 H_{2} O$

Applying Gay Lussac's law,

$3 a + 4 (3 - a) = 10$
$\Rightarrow a = 2$
$∴$ a = 2(Propane) and 3 - 2 = 1 (Butane)
Hence, 2 : 1
(Propane : butane)

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A gas mixture of 3 litres of propane (C3H8) and butane (C4H10) on complete combustion at 25o produced 10 litre CO2. Find out the composition of gas mixture (Propane : Butane)

The correct option is C 2:1 Balanced equation for combustion reactions: C3H8a+5O2→3CO23a+4H2O C4H10(3−a)+132O2→4CO24(3−a)+5H2O Applying Gay Lussac's law, 3a+4(3−a)=10 ⇒a=2 ∴ a = 2(Propane) and 3 - 2 = 1 (Butane) Hence, 2 : 1 (Propane : butane)

A gas mixture of 3 litres of propane $(C_{3} H_{8})$ and butane $(C_{4} H_{10})$ on complete combustion at $25^{o}$ produced 10 litre $C O_{2} .$ Find out the composition of gas mixture (Propane : Butane)