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Question

A sample of gaseous hydrocarbon occupies 0.56L at STP. The complete combustion of the hydrocarbon produces 2.2g of CO2 and 0.9g of H2O. The molecular formula of the compound is?

A
C2H4
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B
C2H6
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C
C3H4
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D
C3H6
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Solution

The correct option is A C2H4
22.4 lit of hydrocarbon= 1 mole
0.5 lit of hydrocarbon= 0.5622.4=0.025 mol
Number of moles of hydrocarbon= 0.025 mol
Number of moles of CO2=massMolecular weight=2.244=0.05 mol
Number of moles of H2O=0.918=0.05 mol

CxHy+(x+y)4CO2xCO2+y2H2O
1 mole of CxHy=x moles of CO2
0.025 moles of CxHy=0.05x moles of CO2
0.025x=0.05
x=2

0.025 mol of CxHx=0.05 mol of H2O
1 mol of CxHx=2 mol of H2O
=y2 mol of H2O
y2=2y=4
Thus compound is C2H4 .

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