What volume of oxygen at STP is required to affect the combustion of 10 litres of propane $(C_{3} H_{8})$ at 540 K and 250 mm of Hg pressure?

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Solution

The correct option is A 8L
$C_{3} H_{8} + 5 O_{8} \to 3 C O_{2} + 4 H_{2} O$
1 Vol 5 Vol

$\begin{matrix} 10 Litres & 50 Litres P_{1} = 250 m m & P_{2} = 760 m m V_{1} = 50 Litres & V_{2} = ? T_{1} = 546 K & T_{2} = 273 K \end{matrix}$

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{250 \times 50}{546} = \frac{760 \times V_{2}}{273} \frac{250 \times 50}{546} = \frac{273}{760} = V_{2}$

$\frac{250 \times 50 \times 273}{546 \times 760} = V_{2} \frac{250 \times 50}{2 \times 760} = V_{2} 8.22 L = V_{2}$

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Q. What volume of oxygen gas at STP is necessary for complete combustion of 20.0 litre of propane

(C H_{3} C H_{2} C H_{3})

, measured at

27^{o} C

and 760 mm of

H g

Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈).

[Atomic mass: C = 12, O = 16, H = 1, Molar volume = 22.4 dm³ at stp]

5.20 litres

3.50 litres

2.58 litres

4.48 litres

1.99 litres

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What volume of oxygen at STP is required to affect the combustion of 10 litres of propane (C3H8) at 540 K and 250 mm of Hg pressure?

The correct option is A 8LC3H8+5O8→3CO2+4H2O 1 Vol 5 Vol 10 Litres50 LitresP1=250 mmP2=760 mmV1=50 LitresV2=?T1=546KT2=273K P1V1T1=P2V2T2250×50546=760×V2273250×50546=273760=V2 250×50×273546×760=V2250×502×760=V28.22L=V2

What volume of oxygen at STP is required to affect the combustion of 10 litres of propane $(C_{3} H_{8})$ at 540 K and 250 mm of Hg pressure?