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Question

Volume of oxygen gas required at NTP to completely oxidise 5.8 g of butane (C4H10) is:

A
14.56 L
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B
11.2 L
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C
22.4 L
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D
10.25 L
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Solution

The correct option is A 14.56 L
C4H10+132O24CO2+5H2O

So , 1 Mole of C4H10 requires 6.5 moles of O2
0.1 Mole of C4H10 requires 0.65 moles O2

So , Volume of oxygen gas required is 0.65×22.4 = 14.56 L

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