# Question bfd77

May 24, 2016

74.8081%, rounded to four decimal places

#### Explanation:

Let the original mixture has no. of grams of Sodium carbonate $= x$
$\therefore$ Number of grams of sodium hydrogen carbonate$= 100 - x$

The given reaction is

$2 \text{NaHCO"_3(s)->"Na"_2"CO"_3(s)+"CO"_2(g)+"H"_2"O} \left(g\right)$

From Mole concept we see that
$2$ Moles of sodium hydrogen carbonate on heating give $1$ mole of sodium carbonate.

Average Molar mass of ${\text{NaHCO}}_{3}$ $= 22.9898 + 1.0079 + 12.011 + 3 \times 15.9994 = 84.0069$
$\therefore$ Average Molar mass of $2$ moles of ${\text{NaHCO}}_{3}$ $= 2 \times 84.0069 = 168.0138$

Average Molar mass of ${\text{Na"_2"CO}}_{3}$ $= 2 \times 22.9898 + 12.011 + 3 \times 15.9994 = 105.9888$

$168.0138$ gm moles of sodium hydrogen carbonate on heating give gm moles of sodium carbonate $= 105.9888$
$100 - x$ -do- $= \frac{105.9888}{168.0138} \times \left(100 - x\right)$
Total sodium carbonate in the final mixture$= \frac{105.9888}{168.0138} \times \left(100 - x\right) + x$, Given $= 90.7$
$\implies \frac{105.9888}{168.0138} \times \left(100 - x\right) + x = 90.7$
Solving for $x$, we get
$x = 74.8081$, rounded to four decimal places
As the original mixtuer is $100$ $g$
Percentage of sodium carbonate=74.8081%, rounded to four decimal places

May 24, 2016

74.8%

#### 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= 100-90.7=9.3g

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 9.3 g will occur for the presence of

$\frac{168 \cdot 9.3}{62} g = 25.2 g \text{ } N a H C {O}_{3}$
in the given mixture

So the amount$N {a}_{2} C {O}_{3} = 100 - 25.2 = 74.8 g$ in 100g mixture

So percentage of Na_2CO_3 =74.8%#