Question

The decomposition of a certain mass of caco3 gave 11.2dm^3 of co2 at stp

Answer 1

its due to gay lussac's law of gaseous volume: any gas has a volume equal to 22.4 litres. so caco3 gives 11.2 gm of co2 at stp

Answer 2

Answer: 50 g

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.

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

According to the given balanced equation:

1 mole of $CaCO_3$ decomposes to give 1 mole of $CO_2$ at STP.

Thus 100 g of $CaCO_3$ decomposes to give 22.4 L or 22.4 $dm^3$ of $CO_2$ at STP       $(1dm^3=1L)$

As 22.4 $dm^3$ of $CO_2$ is produced from 100 g of $CaCO_3$

11.2 $dm^3$ of $CO_2$ is produced from=$\frac{100}{22.4}\times 11.2=50g$ of $CaCO_3$