Answer : 2.24 L / 2.24 dm³
We need to write a equation for the thermal decomposition of Calcium carbonate . The decomposition is as follows ,
[tex]\rm CaCO_3 \xrightarrow{\Delta} CaO + CO_2 [/tex]
The molecular mass of Calcium carbonate is ,
[tex]\rm\longrightarrow Molecular\ mass_{CaCO_3}= 40g + 12g + (16)(3)g = 100g [/tex]
The reaction is already balanced.
- From the reaction , 100g of Calcium carbonate gives 1 mole of carbon dioxide.
[tex]\text{ 100g of Calcium carbonate $\equiv$ 1 mole of carbon dioxide }\\\\\text{ 1g of calcium carbonate $\equiv$ $\rm \dfrac{1}{100}$ mol of carbon dioxide.}\\\\\text{ 10g of calcium carbonate$\equiv$ $\rm \dfrac{1}{100}\times 10= $ 0.1 mol of carbon dioxide.} [/tex]
- And at STP , we know that the volume of 1 mol of gas is 22.4L or 22.4 dm³ .
So that,
[tex]\longrightarrow\rm 0.1\ mole \ occupies \equiv 0.1(22.4L) = 2.24 L [/tex]
Hence ,
[tex]\longrightarrow \underline{\boxed{\bf{\red{ Volume = 2.24 L = 2.24\ dm^3}}}}[/tex]
I hope this helps .
