# How to calculate the percentage composition of a mixture of sodium carbonate/bicarbonate from the volume of carbon dioxide released?

A 1g mixture of $\ce{Na2CO3}$ and $\ce{NaHCO3}$ when heated produce 200 ml of $\ce{CO2}$ at S.T.P. Calculate the percentage composition of the mixture

the thermal decomposition reaction of $\ce{Na_2CO_3}$ takes place after $500~^\circ\mathrm{C}$. But here temperature is not given so to avoid complications I assumed that the heat given is less than that. SO there is thermal Decomposition reaction of $\ce{NaHCO_3}$ which is like this: $$2\ce{NaHCO_3 ->Na_2CO_3 + H_2O + CO_2}$$ So we have got 200ml of $\ce{CO_2}$ and we know that $1~\mathrm{mol} = 22.4~\mathrm{L}$ so $200~\mathrm{ml} = 0.2~\mathrm{L}= 1/112~\mathrm{mol}$ of $\ce{CO_2}$.

From stoichiometric calculations: $2~\mathrm{mol}$ of $\ce{NaHCO_3}$ give $1~\mathrm{mol}$ of $\ce{CO_2}$ So, $1/112~\mathrm{mol}$ of $\ce{CO_2}$ need $2\cdot(1/112)=1/56~\mathrm{mol}$ of $\ce{NaHCO_3}$

Let the weight of $\ce{Na_2CO_3}$ be $x~\mathrm{g}$ so the weight of $\ce{NaHCO_3}$ will be $(1-x)~\mathrm{g}$.

molar mass of $\ce{NaHCO_3}= 23 + 1 + 12 + 16 \cdot 3 = 84~\mathrm{g}$ but weight of $\ce{NaHCO_3}= 84 \cdot 1/56 = 84/56~\mathrm{g}$, but it roughly equals to 1 point something, but the weight of the original mixture is 1 gram. So what and where is the problem ?

• actually my teacher latter corrected the question from 200ml to 20ml – anni May 19 '15 at 22:44

## 1 Answer

I feel there is some error in the question itself.
I can't see any error in your calculations.
The amount of initial reactants can't be 1 gram.

By the way,

But here temperature is not given so to avoid complications I assumed that the heat given is less than that.

its mentioned STP.