Question

Calculate the volume of carbon dioxide at ntp evolved by strong heating of $20g$ calcium carbonate?

Answer 1

Ntp is used for gaseous state normal value of temperature is taken as $293.15K$ and pressure at $1atm$ is used in calculations of moles or volume for ideal gases.

Complete step by step answer:
Firstly, balance the equation in which calcium carbonate decomposes on heating to from calcium oxide and carbon dioxide
$CaC{O_3} \to CaO + C{O_2} \uparrow$
Here $1$mole of $CaC{O_3}$ is equal to $100g$ and $1$ mole of $CaO$ is equal to $22.4l$ at ntp.
The weight of calcium carbonate heated $= 20g$
From the equation we see that $1$ mole of calcium carbonate revolves one mole of carbon dioxide. Now let us calculate the molar mass of calcium carbonate $\left( {CaC{O_3}} \right)$ which can be calculated as-
Atomic number of calcium is $20$ and atomic mass of calcium is $40$
Atomic number of carbon is $6$ and atomic mass of carbon is $12$
Atomic number of oxygen is $8$ and atomic mass of oxygen is $16$
Now molar mass of CaC{O_3}$$$$ = 40 + 12 + 16 + 3 i.e. $\;100g$.
According to the balance reaction $100g$ of $CaC{O_3}$ produce $1$ mole of $C{O_2}$ that is $22.4l$ at ntp.
Now we have the given weight of calcium carbonate which is $20g$. After balancing the equation for heating of $CaC{O_3}$ we get
$CaC{O_3} \to CaO + C{O_2} \uparrow$
We know that one mole of $CaC{O_3}$ weighs $100g$ . Therefore the number of moles present in $20g$ of calcium carbonate is equal to $\dfrac{{20}}{{100}}$ which is equal to $0.2$ moles.
Now the number of moles of $C{O_2}$ given by $0.2$ moles of calcium carbonate is equal to $0.2$ moles.
Now according to Avogadro law for gases $1$ mole of any gas at normal temperature and pressure occupies $22.4l$ of volume.
Volume occupied by $0.2$ of carbon dioxide $= 0.2 \times 22.4l$
$\Rightarrow$ Volume occupied by $0.2$ of carbon dioxide $= 4.48l$

Hence the volume of carbon dioxide at ntp evolved by strong heating of $20g$ of calcium carbonate is $4.48l$ .

Note: We can solve this question b using this alternate method-
$100g$ of calcium carbonate produces one mole of $C{O_2}$ is equal to $22.4l$ at ntp
So 20g$$$$CaC{O_3} will produce $= \dfrac{{22.4}}{{100}} \times 20$
\Rightarrow $$$20g$$$$CaC{O_3}$$ will produce = 4.48l$
We have to remember that heating calcium carbonate leaves behind white Residue. A colourless and odourless gas is evolved which turns lime water milky.