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Stoichiometric Calculations

A 1.2 g ...

Question

A $1.2 g$ mixture of $N a_{2} C O_{3}$ and $K_{2} C O_{3}$ was dissolved in water in form $100 c m^{3}$ of a solution. $20 c m^{3}$ of this solution required $40 c m^{3}$ of $0.1 N H C l$ for neutralisation. Calculate the mass of $N a_{2} C O_{3}$ and $K_{2} C O_{3}$ in the mixture.

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Solution

The total weight of mixture is 1.20 g.
Volume of the solution = 100mL
Volume the solution taken for neutralisation $V_{2}$ = 20mL
Volume of HCl solution $V_{1} = 40 c m^{3}$ = 40mL
Normality of HCl $N_{1}$ = 0.1N

Now. for calculating the mass of $N a_{2} C O_{3}$

$N_{1} \times V_{1} = N_{2} \times V_{2}$
$0.1 \times 40 = N_{2} \times 20$
$N_{2}$ = 0.2N

It means that 1L of solution contains 0.2 gram equivalent.
$∴$ 0.1L of solution contains 0.02 gram equivalent.

In a mixture, number of gram equivalent = sum of g equivalent of ( $K_{2} C O_{3} + N a_{2} C O_{3}$ )

We know that Number of gram equivalent = $\frac{m a s s}{equivalent mass}$
and
Equivalent mass = $\frac{molar mass}{Valency Factor}$

Equivalent mass of $N a_{2} C O_{3}$ = $\frac{106}{2} = 53 g$

Equivalent mass of $K_{2} C O_{3}$ = $\frac{138}{2} = 69 g$

Let weightof $N a_{2} C O_{3} = x g m .$
weight of $K_{2} C O_{3} = (1.2 - x) g m$

0.02 = $\frac{x}{53} + \frac{1.2 - x}{69}$
On calculating we get , x= 0.59gm = 0.6 gm

Weight of $N a_{2} C O_{3} = 0.6 g m .$
Weight of $K_{2} C O_{3} = 0.6 g m$

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