Question

Calculate the volume of $0.1$ M $HCl$ required to react completely with $1$ g of a equimolar mixture of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$.

• A. 15.78 mL
• B. 187.8 mL
• C. 200 mL
• D. 157.8 mL

Answer 1

The correct option is D 157.8 mL

Let, $x$ moles each of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ are present in $1$ g of the mixture.
Molar mass of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ are $106$ g/mol and $84$ g/mol respectively.
Thus, according to the question,
$1\text{g}=x\text{mol}\times 106\text{g/mol}+x\text{mol}\times 84\text{g/mol}$
On solving, $x=5.26\times {10}^{-3}mol$
Let, volume of HCl required be $V$ mL
So, number of moles present in $V$ mL of $0.1$ M $HCl$ $=0.1\times V\times {10}^{-3}\text{mol}$
Reaction of $HCl$ with $N{a}_{2}C{O}_3$
$N{a}_{2}C{O}_3+2HCl\to 2NaCl+C{O}_2+H_{2}O$
Thus, $1$ mole of $N{a}_{2}C{O}_3$ requires $2$ mole of $HCl$.
Hence $x$ moles of $N{a}_{2}C{O}_3$ will require $2x$ moles of $HCl$
Again, reaction of $HCl$ with $NaHC{O}_3$
$NaHC{O}_3+HCl\to NaCl+C{O}_2+H_{2}O$
So, $1$ mole of $NaHC{O}_3$ requires $1$ mole of $HCl$
Hence, $x$ mole of $NaHC{O}_3$ requires $x$ mole of $HCl$
So, $0.1\times V\times {10}^{-3}=x+2x=3\times 5.26\times {10}^{-3}$
On solving, $V=157.8\text{mL}$