Question

$V$ mL of $0.1$ N $HCl$ is required to react completely with $1$ g mixture of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ containing equimolar amounts of two. $V$ is:

• A. $157.89$
• B. $257.89$
• C. $152.89$
• D. none of the above

Answer 1

Answer: (A)

$157.89$

The molar masses of $N{a}_{2}C{O}_3$ and $NaHC{O}_3$ are 106 g/mol and 84 g/mol respectively.

Let x moles of $N{a}_{2}C{O}_3$ are present in 1 g of mixture.

Mass of $N{a}_{2}C{O}_3$ present in the mixture $=106x$ g

Mass of $NaHC{O}_3$ present in the mixture $=1-106x$

Moles of $NaHC{O}_3$ present in the mixture $=\frac{1-106x}{84}$ moles

The mixture contains equimolar amounts of two.

$\frac{1-106x}{84}=x$

$1-106x=84x$

$1=190x$

$x=0.00526$ moles

Moles of $HCl$ required for neutralization $=3\times 0.00526=0.015789$ moles

Volume of $0.1NHCl$ required$=\frac{0.015789moles}{0.1moles/L}=0.15789L=157.89mL$