Question

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

Answer 1

Hint: 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.