Question

Assuming fully decomposed, the volume of $CO_2$ released at STP on heating 9.85 g of $BaCO_3$ (at. mass of Ba = 137) will be

• A. 1.12 L
• B. 0.84 L
• C. 2.24 L
• D. 4.96 L

Answer 1

Answer: (A)

1.12 L

Answer 2

The correct answer is A:1.12 L
On decomposition, $BaCO_3$ liberates $CO_2$ as
$_{197 g}^{BaCO_3} \, \, \, \, \, \rightarrow BaO +_{22.4 L \, at \, STP }^{ \, \, \, \, \, \, CO_2 \uparrow}$
$\therefore \, \, 197 g \, of \, BaCO_3 \, gives \, = 22.4 L \, of\, CO_2 \, at \, STP$
$\therefore 9.85 \, g \, of \, BaCO_3 \, will \, give\, = \frac {22.4 \times 9.85}{197} = 1.12 L$