Question

Calculate the volume of co2 formed at ntp by heating 4.2g of nahco3

Answer 1

2NaHCO3=Na2CO3+CO2+H2O
i.e2moles of nahco3 gives 1mole of co2
Then 0.05 moles of nahco3 gives ?co2
i.e.0.025 moles
Now for volume 0.025×22.4=0.56l

Answer 2

Answer: The volume of $CO_2$ formed will be 0.56 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 $NaHCO_3$ = 4.2 g

Molar mass of $NaHCO_3$ = 84 g/mol

Putting values in above equation, we get:

$\text{Moles of }NaHCO_3=\frac{4.2g}{84g/mol}=0.05mol$

The chemical reaction for the heating of sodium hydrogen carbonate follows the equation:

$2NaHCO_3\rightarrow Na_2CO_3+CO_2+H_2O$

By Stoichiometry of the reaction:

2 moles of sodium bicarbonate produces 1 mole of carbon dioxide.

So, 0.05 moles of sodium bicarbonate will produce = $\frac{1}{2}\times 0.05=0.025mol$ of carbon dioxide.

At STP:

1 mole of a gas occupies 22.4 liters of volume.

So, 0.025 moles of carbon dioxide will occupy $\frac{22.4}{1}\times 0.025=0.56L$ of volume.

Thus, the volume of $CO_2$ formed will be 0.56 L.