Question

A 2 g sample containing $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ loses 0.62 g when heated to $300{}^{o}C,$ the temperature at which $NaHC{O}_3$ decomposes to $N{a}_{2}C{O}_3,C{O}_2$ and $H_{2}O.$ What is the percentage of $N{a}_{2}C{O}_3$ in the given mixture.

Answer 1

On heating $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ , $N{a}_{2}C{O}_3$ remains unchanged while $NaHC{O}_3$ changes into $N{a}_{2}C{O}_3$, $C{O}_2$ and $H_{2}O.$
The loss in weight is due to removal of $C{O}_2$ and $H_{2}O$ which escape out on heating.
$\therefore massofN{a}_{2}C{O}_3$ in the product
$=2.00-0.62=1.38g.$
Let, the weight of $N{a}_{2}C{O}_3$ in the mixture be X g.
$\therefore$ wt. of $NaHC{O}_3=(2.00-X)g.$
Since, $N{a}_{2}C{O}_3$ in the products contains X g of unchanged reactant $N{a}_{2}C{O}_3$ and rest produced from $NaHC{O}_3.$
Then, wt. of $N{a}_{2}C{O}_3$ produced by $NaHC{O}_3$ only $=(1.38-X)g.$
Now, we have,
$\underset{(2.0-X)g}{NaHC{O}_3}\to \underset{(1.38-X)g}{N{a}_{2}C{O}_3+}(H_{2}O+C{O}_2)\uparrow$
Applying POAC for Na atoms, moles of Na in $NaHC{O}_3$ = moles of Na in $N{a}_{2}C{O}_3$
$1\times molesofNaHC{O}_3=2\times molesofN{a}_{2}C{O}_3$
$\frac{2.0-X}{84}=2\times \frac{1.38-X}{106}\left[\begin{array}{c}\text{molar mass of}NaHC{O}_3=84\\ \text{molar mass of}N{a}_{2}C{O}_3=106\end{array}\right]$
$X=0.32g.$
$\therefore$ % of $N{a}_{2}C{O}_3=\frac{0.32}{2.0}\times 100=16$%.