Volume of $C O_{2}$ obtained at STP by the complete decomposition of 9.85 gm $B a C O_{3}$ is:

[Molecular weight of $B a C O_{3} = 197$ ]

2.24 litre

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1.12 litre

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0.85 litre

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0.56 litre

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Solution

The correct option is B 1.12 litre
Volume of 1 mole of $C O_{2}$ at STP = 22.4 litre
Volume of $C O_{2}$ obtained after complete decomposition of 9.85gm of $B a C O_{3}$ is:

$B a C O_{3} \to B a O + C O_{2}$
No of moles of $B a C O_{3}$ obtained = $\frac{9.85}{197}$ = 0.05 mole

As 1 mole of $B a C O_{3}$ gives 1 mole of $C O_{2}$ ,

Therefore, number of moles of $C O_{2}$ obtained = 0.05 mole

$∴$ Volume of $C O_{2}$ obtained at STP $= 22.4 \times 0.05 = 1.12$ litre

Hence, option $B$ is correct.

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