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,
$1g=xmol\times 106g/mol+xmol\times 84g/mol$
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 $HCl$ $=0.1\times V\times {10}^{-3}mol$
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 mole of $N{a}_{2}C{O}_3$ will require 2x mole 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 mL