Question

How many mL of $0.1NHCl$ are required to react completely with $1g$ mixture of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ containing equimolar amounts of both?

• A. $200.7$
• B. $300.0$
• C. $257.8$
• D. $157.8$

Answer 1

The correct option is D $157.8$

Let $xmol$ of each react completely
$\Rightarrow \text{weight of}N{a}_{2}C{O}_3=106\times xg\text{and}\text{weight of}NaHC{O}_3=84\times xg$
$\Rightarrow 106x+84x=1\Rightarrow x=\frac{1}{190}mol$

One mol of sodium carbonate will react with 2 mol of HCl and 1 mol of sodium bicarbonate will react with 1 mol of HCl.

$\text{Equivalent of}N{a}_{2}C{O}_3+\text{Equivalent of}NaHC{O}_3=\text{Equivalent of}HCl$
$\Rightarrow \frac{1}{190}\times 2+\frac{1}{190}\times 1=0.1\times \frac{V}{1000}$
$\Rightarrow \frac{3}{190}=\frac{0.1\times V}{1000}\Rightarrow V=157.8mL$