Question

A sample of $CaC{O}_3$ and $MgC{O}_3$ weighing $2.21$ g is ignited to a constant mass of $1.152$ g. The volume (in ml) of $C{O}_2$ evolved at $0^{0}C$ and $76$ cm pressure is (as the nearest integer) _______.

Answer 1

Let the mixture contain $a$ g $CaC{O}_3$ and $b$ g $MgC{O}_3$.

$\therefore a+b=2.21...\left(i\right)$

On heating $CaC{O}_3\to CaO+C{O}_2$

$MgC{O}_3\to MgO+C{O}_2$

Moles of $CaO$ $=$ Moles of $CaC{O}_3=$ $\frac{a}{100}$.

Mass of $CaO$ $=\left(\frac{a}{100}\right)\times 56g$.

Similarly, mass of MgO $=\left(b/84\right)\times 40g$.

Therefore, $\frac{56a}{100}+\frac{40b}{84}=1.152...\left(ii\right)$ (i.e., mass of residue left)

Solving Eqs. (i) and (ii), $a=1.19g$; $b=1.02g$

Also , moles of $C{O}_2$$=$ mole of $CaC{O}_3$+ mole of $MgC{O}_3$

$=\frac{1.19}{100}+\frac{1.02}{84}=0.0241$.

Volume of $C{O}_2$ at NTP $=0.0241\times 22400$ $=539.8$ mL.

So, the answer is $540$ mL.