Question

Volume of co2 obtained at stp by complete decomposition of 9.85gm baCo3 is

Answer 1

:
(1) 2.24 lit.
(2)1.12 lit.
(3) 0.84 lit.
(4) 0.56 lit.

Answer 2

Answer:  1.12L of $CO_2$ by complete decomposition of 9.85 g of $BaCO_3$.

Explanation: According to Avogadro's law, 1 mole of every gas occupy 22.4 L and weighs equal to the molecular mass of that substance.

According to the given balanced equation:

$BaCO_3(s)\rightarrow BaO(s)+CO_2(g)$

1 mole of $BaCO_3$ decomposes to give 1 mole of $CO_2$ at STP.

Thus 197 g of $BaCO_3$ decomposes to give 22.4 L of $CO_2$ at STP.          

9.85 g $BaCO_3$ decomposes to give=$\frac{22.4}{197}\times 9.85g=1.12L$ of $CO_2$ at STP .