Question

Volume of CO2 obtained at STP by the complete decomposition of 9.85g Na2CO3 is

Answer 1

on calculating this reaction we might get answer 2.0 but as Na2Co3 ( sodium carbonate) is soda ash and it does not decompose on heating and CO2 does not involve.
so we get 0 litre volume.

I hope u understand
thanks

Answer 2

Answer: The volume of $CO_2$ obtained will be 240.8 L.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium carbonate = 9.58 g

Molar mass of sodium carbonate = 106 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium carbonate}=\frac{9.85g}{106g/mol}=0.093mol$

Equation for the decomposition of sodium carbonate follows:

$Na_2CO_3\rightarrow Na_2O+CO_2$

By Stoichiometry of the reaction:

1 mole of sodium carbonate produces 1 mole of carbon dioxide.

So, 0.093 moles of sodium carbonate will produce = $\frac{1}{1}\times 0.093=10.75mol$ of carbon dioxide.

At STP:

1 mole of a gas occupies 22.4 L of volume.

So, 10.75 moles of carbon dioxide will occupy $10.75\times 22.4=240.8L$ of volume.

Hence, the volume of $CO_2$ obtained will be 240.8 L.