$11.2 L$ of gas at STP weight 14g. The gas would be :

H_{2}

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C O

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B_{2} H_{6}

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All of the above

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Solution

The correct options are
C $B_{2} H_{6}$
D $C O$
No. Of moles of that gas $= \frac{11.2 L}{22.4 L} = \frac{1}{2} mole$

Given mass =14g

We know,
No. Of moles =given mass /molecular mass

$\frac{1}{2} mole = \frac{14}{molar mass}$

molar mass=28g

the molar mass of $C O$ is 28

also, the molar mass of $B_{2} H_{6}$ is 28g

Thus, the gas could be CO or B $_{2}$ H $_{6}$

Hence, the correct options are $B$ and $C$

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