On the thermal decomposition of 84 grams of sodium bicarbonate, the volume of $C O_{2}$ liberated at STP is:

11.2 L

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22.4 L

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5.6 L

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44.8 L

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Solution

The correct option is C 11.2 L
$2 N a H C O_{3} \to N a_{2} C O_{3} + C O_{2} + H_{2} O$

2 moles of sodium bicarbonate on thermal decomposition gives 1 mole of $C O_{2}$ .

Hence, thermal decomposition of $1$ mole of sodium bicarbonate ( $84 g m$ ) will give $0.5$ moles of $C O_{2}$ .

At STP, 1 mole of $C O_{2}$ occupies a volume of $22.4 L$ .

Hence, 0.5 moles of $C O_{2}$ will occupy a volume of $\frac{22.4}{2} = 11.2 L$ .

Hence, option A is correct.

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What will be the volume (in litres) of oxygen liberated (at STP) when 17 grams of $H_{2}$ $O_{2}$ decomposed on heating?

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C O_{2} (g)

2 N a H C O_{3} Δ - \to N a_{2} C O_{3} + C O_{2} + H_{2} O

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