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Volume of Gases and Number of Moles

If 12.6 g of ...

Question

If 12.6 g of $N a H C O_{3}$ is added to 20.0 g of $H C l$ solution, the residue solution is found to weigh 24.0 g. What is the mass and volume of $C O_{2}$ released at NTP in the reaction?

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Solution

The reaction is given as $N a H C O_{3} + H C l ⟶ N a C l + C O_{2} ↑ + H_{2} O$
The loss in mass is due to the liberation of $C O_{2} .$
Total mass reactants $= 12.6 + 20.0 = 32.6 g$
Mass lost $= 32.6 - 24.0 = 8.6 g = m a s s o f C O_{2}$
Number of moles of $C O_{2} = \frac{8.6}{44} = 0.195$
$∴ v o l u m e = 0.195 \times 22.4 = 4.37 d m^{3}$

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