How many grams of sodium carbonate is obtained on heating $8.4 g$ of $N a H C O_{3} (s)$ ?

2.1 g

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5.3 g

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4.2 g

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4.6 g

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Solution

The correct option is B $5.3 g$
$2 N a H C O_{3} Heat - - \to N a_{2} C O_{3} + C O_{2} + H_{2} O$
Given mass of $N a H C O_{3}$ = $42 g$
Molar mass of $N a H C O_{3}$ = $84 g$
Number of moles of $N a H C O_{3}$ = $\frac{8.4}{84} = 0.1 moles$
$2 moles$ of $N a H C O_{3}$ gives $1 mole$ of $N a_{2} C O_{3}$ on heating.
Number of moles of $N a_{2} C O_{3}$ produced on heating $0.1 moles$ of $N a H C O_{3}$ = $\frac{0.1}{2} = 0.05 mol$
Mass of $N a_{2} C O_{3}$ produced
$= Number of moles of N a_{2} C O_{3} produced \times Molar mass of N a_{2} C O_{3}$
= $0.05 moles \times 106 {g mol}^{- 1}$
= $5.3 g$

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