Question

What is the volume of $0.1MHCl$ required to react completely with $1g$ mixture of equimolar amount of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$?

• A. $106mL$
• B. $128mL$
• C. $156mL$
• D. $212mL$

Answer 1

The correct option is C $156mL$

$\underset{n_f=2}{N{a}_{2}C{O}_3}+2HCl\to 2NaCl+N{a}_{2}C{O}_3$
$\underset{n_f=1}{NaHC{O}_3}+HCl\to NaCl+C{O}_2+H_{2}O$

Let $xg$ of $N{a}_{2}C{O}_3$ and $(1-x)g$ of $NaHC{O}_3$ are present in the mixture.

$\text{Equivalents of}N{a}_{2}C{O}_3+\text{Equivalents of}NaHC{O}_3=\text{Equivalents of}HCl$
$\therefore \frac{x}{106}\times 2+\frac{1-x}{86}\times 1=0.1\times V...\left(1\right)$

Given, $\text{Moles of}N{a}_{2}C{O}_3=\text{Moles of}NaHC{O}_3$
$\therefore \frac{x}{106}=\frac{1-x}{86}...\left(2\right)$
$86x=106-106x$
$\therefore x=0.552$

From equation $\left(1\right)$,
$\frac{0.552}{106}\times 2+\frac{1-0.552}{86}\times 1=0.1\times V$
$0.0104+0.0052=0.1\times V$
$0.0156=0.1\times V$
$\therefore V=0.156L$
or, $V=156mL$