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Preparations and Uses of Sodium Carbonate

When 16.8 g...

Question

When $16.8 g$ of white solid $X$ was heated $4.4 g$ of acid gas : $(A)$ that turned lime water milky was driven of together with $1.8 g$ of a gas $(B)$ which condensed to a colourless liquid the solid that remained $(Y)$ dissolved in water to give an alkaline solution, which with excess of $B a C l_{2}$ solution gave a white precipitate $(Z)$ . The precipitate effervescence with acid giving of $C O_{2}$ gas. Identify the compound $A, B$ and write the chemical equations for the thermal decomposition of $X$ .

(A) - C O_{2}

(B) - H_{2} O X \to N a_{2} C O_{3}

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(A) - C O_{2}

(B) - H_{2} O_{2} X \to N a C O_{3}

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(A) - C O_{2}

(B) - H_{2} O_{2} X \to N a_{2} C O_{3}

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(A) - C O_{2}

(B) - H_{2} O X \to N a C O_{3}

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Solution

The correct option is A $(A) - C O_{2}$ $(B) - H_{2} O X \to N a_{2} C O_{3}$
The given facts
$X (s) H e a t \to A (g) + B (g) + Y (s)$
$Y w a t e r \to A l k a l i n e s o l u t i o n B a C l_{2} \to Z (W h i t e p p t)^{+} C O_{2}$
Since $Z$ on reaction with acid evolves $C O_{2}$ it must be $B a C O_{3}$ . As $B a C O_{3}$ is obtained by reaction of $B a C l_{2}$ with alkaline solution, $Y$ must be a carbonate of alkali metal.
Gas $A$ must be $C O_{2}$ which turns lime water milky due to formation of $C a C O_{3}$ . With this $X$ must be bicarbonate of alkali metal. Which on heating produce
$2 M H C O_{3} \to C O_{2} + H_{2} O + M_{2} C O_{3}$

By calculating from the given data the molar mass of $M$
$\Rightarrow 2 (M + 61) g$ of $M H C O_{3}$ gives $44 g C O_{2}$
$\Rightarrow \frac{2 (M + 61)}{44 g C O_{2}} \times 4.4 g C O_{2} = 16.8 g M H C O_{3}$
$\Rightarrow M = 23 g m o l^{-}$
Therefore $M$ must be $N a$ .
$A = C O_{2}$
$B = H_{2} O$
$X = N a_{2} C O_{3}$
$N a_{2} C O_{3} \to C O_{2} + H_{2} O + N a H C O_{3}$

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