# Question f137a

Jul 9, 2016

=79%

#### Explanation:

On heating the given mixture of 100g of $N a H C {O}_{3} \mathmr{and} N {a}_{2} C {O}_{3}$ ,only$N a H C {O}_{3}$ will decompose according to the following balanced equation but other component $N {a}_{2} C {O}_{3}$ will not.

The Equation

$2 N a H C {O}_{3} \left(s\right) \to N {a}_{2} C {O}_{3} \left(s\right) + C {O}_{2} \left(g\right) + {H}_{2} O \left(g\right)$

According to this balanced equation 2 moles of solid $N a H C {O}_{3}$ produces 1 mole of ${H}_{2} O \left(g\right)$and 1 mole of $C {O}_{2} \left(g\right)$ as volatile matter

Molar mass of
$N a H C {O}_{3} = 23 + 1 + 12 + 3 \cdot 16 = 84 \frac{g}{\text{mol}}$

Molar mass of
$C {O}_{2} = 12 + 2 \cdot 16 = 44 \frac{g}{\text{mol}}$

Molar mass of ${H}_{2} O = 2 \cdot 1 + 16 = 18 \frac{g}{\text{mol}}$

By the problem the decrease in mass = initial mass -residual mass= =0.31g

According to the decomposition reaction total 44+18=62g decrease in mass occurs for presence of $2 \cdot 84 g = 168 g$ of $N a H C {O}_{3}$ in the mixture.

So decrease in mass of 0.31 g will occur for the presence of

$\frac{168 \cdot 0.31}{62} g = 0.84 g \text{ } N a H C {O}_{3}$
in the given mixture

So the amount$N {a}_{2} C {O}_{3} = 4 - 0.84 = 3.16 g$ in 4g mixture

So percentage of Na_2CO_3 =3.16/4xx100%=79%#