Question

volume of CO2 obtained at STP by the complete decomposition of 9.85gm of BaCO3 is
1) 2.24l
2) 1.12l
3)0.85l
4)0.56l

Answer 1

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Answer 2

Answer: 2) 1.12l

Explanation:

Moles of $BaCO_3$ =$\frac{\text{ given mass }}{\text{ molar mass}}$

Moles of $BaCO_3$ =$\frac{9.85g}{197g/mol}=0.05moles$

$BaCO_3\rightarrow BaO+CO_2$

According to Avogadro's law, 1 mole of any gas occupies 22.4 L of volume at STP.

1 mole of  $BaCO_3$ produces 22.4 L of $CO_2$ at STP

0.05  mole of  $BaCO_3$ produces=$\frac{22.4}{1}\times 0.05=1.12L$ of $CO_2$ at STP