An impure sample of calcium carbonate contains 80 % pure calcium carbonate. 25g of the impure sample reacted with excess hydrochloric acid. Calculate the volume of carbon dioxide at NTP obtained from this sample.

Hint: The reaction taking place when calcium carbonate reacts with HCl is as-
$CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O$
Further, 100g of impure sample contains 80 g of pure calcium carbonate. So, the 25% which is reacting is an impure sample. This means the pure sample will be less than 25 g. First we will find the pure sample present in 25 g and then find out the volume.

Complete step by step answer:
In the question, we have been given that an impure sample of calcium carbonate contains 80% pure calcium carbonate. So, this means if we have 100 g of impure sample of calcium carbonate then it will contain 80 g of calcium carbonate in it.
The next thing given is that 25 g of sample reacts with HCl acid.
Let us first write the equation -

$CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O$

We can see in the equation that carbon dioxide is formed as a side product to calcium chloride along with water. We have to calculate the volume of this carbon dioxide produced at NTP when 25g of sample reacts with HCl acid.
So, let us proceed step by step.
We have seen above that 100g of impure sample of calcium carbonate will contain 80g of calcium carbonate in it.
Thus, 25 g of impure sample of calcium carbonate will contain = $\dfrac{{80}}{{100}} \times 25$
25 g of impure sample of calcium carbonate = 20 g of pure calcium carbonate
As we know, the molecular mass of calcium carbonate is 100 g.
Thus, 100 g or 1 mol of pure calcium carbonate will result in formation of 22.4 L of carbon dioxide.
$CaC{O_3} + 2HCl \to CaC{l_2} + C{O_2} + {H_2}O$
 1mol $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$ 22.4 litre
 100g
But we have 20 g of pure calcium carbonate in 25 g of impure sample.

So, 20g of pure Calcium carbonate will release = $\dfrac{{22.4}}{{100}} \times 20$ = 4.48 L of $C{O_2}$
So, 4.48 L of $C{O_2}$ is our answer.


Note: It must be noted that 1 mole of any gas is equal to the 22.4 L of the gas at NTP. The NTP stands for Normal Temperature and Pressure. At NTP, the temperature is taken to be 293.15K and the pressure is 1 atm.
As from the reaction, it is clear that 1 mole of calcium carbonate will form 1 mole of carbon dioxide and further, 1 mole of carbon dioxide gas at NTP will be equal to 22.4 L of gas.
So, we put 1 mole of calcium carbonate producing 22.4 L of carbon dioxide.