Question

Write an equation for the thermal decomposition of CaCO3. Determine the volume of CO2 measured at s t p. that would be produced by the thermal decomposition of 10g of CaCO3. [Ca= 40, O =16 C= 12, molar volume = 22.4dm3 at STP]​

Answer 1

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 ,

$\rm CaCO_3 \xrightarrow{\Delta} CaO + CO_2$

The molecular mass of Calcium carbonate is ,

$\rm\longrightarrow Molecular\ mass_{CaCO_3}= 40g + 12g + (16)(3)g = 100g$

The reaction is already balanced.

• From the reaction , 100g of Calcium carbonate gives 1 mole of carbon dioxide.

$\text{ 100g of Calcium carbonate \equiv 1 mole of carbon dioxide }\\\\\text{ 1g of calcium carbonate \equiv \rm \dfrac{1}{100} mol of carbon dioxide.}\\\\\text{ 10g of calcium carbonate \equiv \rm \dfrac{1}{100}\times 10= 0.1 mol of carbon dioxide.}$

• And at STP , we know that the volume of 1 mol of gas is 22.4L or 22.4 dm³ .

So that,

$\longrightarrow\rm 0.1\ mole \ occupies \equiv 0.1(22.4L) = 2.24 L$

Hence ,

$\longrightarrow \underline{\boxed{\bf{\red{ Volume = 2.24 L = 2.24\ dm^3}}}}$

I hope this helps .