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Question

A sample of gaseous hydrocarbon occupying 1.12 litre at NTP, when completely burnt in air produced 2.2 g CO2 and 1.8 g H2O. Calculate the mass of hydrocarbon taken and the volume of O2 at NTP required for its combustion.


A

1.14 L

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B

2.24 L

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C

2.4 L

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D

2.8 L

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Solution

The correct option is B

2.24 L


You know that a hydrocarbon is made up of C and H

CaHb + (a + b4)O2 a CO2 + b2 H2O (I) is the combustion of the hydrocarbon.

1 mole of CaHb = (44a) g of CO2

1.12 L of CaHb = (1.12×44a22.4) g of CO2

From question 1.12×44a22.4 = 2.2g a = 1

Again, 22.4 L of CaHb = (b2 × 18)g of H2O

1.12 of CaHb = 1.12×b2×1822.4

From question, 1.12×b2×1822.4 = 1.8 b = 4

The hydrocarbon of methane (CH4)

Number of moles of oxygen = ( 1 + 44) = 2 moles.

From reaction,

22.4 L of CH4 = 2 × 22.4 L of O2

1.12 L of CH4 = 1.12×2×22.422.4

1.12 L of CH4 = 2.24L of O2


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