The volume of oxygen required for the complete combustion of 8.8 g of propane $(C_{3} H_{8})$ is:
[ $C = 12, O = 16, H = 1,$ Molar Volume $= 22.4$ dm $^{3}$ at S.T.P]

22.4 L

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44.8 L

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11.2 L

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8.8 L

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Solution

The correct option is A 22.4 L
Given that
Mass of propane $= 8.8 g$
Number of moles $= \frac{m a s s}{m o l a r m a s s} = \frac{8.8}{44} = 0.2 m o l$

Now,

$C_{3} H_{8}$ + $5 O_{2}$ $⟶$ $3 C O_{2}$ + $4 H_{2} O$

$1$ mole of propane requires $5$ $\times$ { $22.4$ } litre $O_{2}$ .

So, $0.2$ mole propane requires $5$ $\times$ { $22.4$ } $\times$ { $0.2$ } $=$ $22.4$ litre $O_{2}$

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Similar questions

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

(i) Calculate the volume of 320 g of SO₂ at stp (Atomic mass: S = 32 and O = 16)
(ii) State Gay-Lussac's Law of combining volumes.
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C₃H₈). (Atomic mass: C = 14, O = 16, H = 1, Molar Volume = 22.4 dm³ at stp)

(a)

(i) Calculate the volume of 320 of $S O_{2}$ at STP. (Atomic mass : S = 32 and O = 16).
(ii) State Gay-Lussac's law of combining volume
(iii) Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane $(C_{3} H_{8})$ (Atomic mass; C = 12, 0 = 16, H = 1, Molar volume $- 22.4 d m^{3}$

(b)

(i) An organic compound with vapour density = 94 contains
C = 12.67% H = 2.13 % and Br = 85.11% Find the molecular formula.
[Atomic mass ; C = 12, H = 1, Br = 80]
(ii) Calculate the mass of
(1) $10^{22}$ Atoms of sulphur.
(2) 0.1 mole of carbon dioxide.
[Atomic mass; S = 32, C = 12 and O = 16 and Avogadro's number $= 6 \times 10^{23}$

Q. How much volume of

C O_{2}

S . T . P

is liberated by the combustion of

100 c m^{3}

of propane

(C_{3} H_{8})

12.6 g

nonene

C_{9} H_{18}

is combusted in the presence of air. The percentage of oxygen in air is 21 percent by volume at STP. The volume of air needed for complete combustion of nonene is

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The volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8) is: [C=12,O=16,H=1, Molar Volume =22.4 dm3 at S.T.P]

The correct option is A 22.4 LGiven that Mass of propane =8.8 gNumber of moles =massmolar mass=8.844=0.2 molNow,C3H8 + 5O2 ⟶ 3CO2 + 4H2O1 mole of propane requires 5×{22.4} litre O2.So, 0.2 mole propane requires 5×{22.4}×{0.2} = 22.4 litre O2

The volume of oxygen required for the complete combustion of 8.8 g of propane $(C_{3} H_{8})$ is:
[ $C = 12, O = 16, H = 1,$ Molar Volume $= 22.4$ dm $^{3}$ at S.T.P]