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Stoichiometric Calculations

Three mol of ...

Question

Three mol of $N a_{2} C O_{3}$ is allowed to react with a solution containing 6 mol of $H C l$ . Find the volume of $C O_{2}$ gas (in L) produced at STP.
$N a_{2} C O_{3} + 2 H C l ⟶ 2 N a C l + C O_{2} + H_{2} O$

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Solution

$\begin{matrix} From the reaction: & N a_{2} C O_{3} & + 2 H C l & \to 2 N a C l + C O_{2} + H_{2} O Given mol: & 3 mol & 6 mol Given mol ratio: & 1 & 2 Stoichiometric coefficients ratio: & 1 & 2 \end{matrix}$

Since given mol ratio of reactants is the same as stoichiometric coefficient ratio, therefore no reactant will be left over.

Now use mol - mol analysis to calculate mol, and hence, the volume of $C O_{2}$ produced at STP is,
$\frac{Moles of N a_{2} C O_{3}}{1} = \frac{Moles of C O_{2} produced}{1}$
Moles of $C O_{2}$ produced = 3 mol
Volume of $C O_{2}$ produced at STP = $3 \times 22.4 L = 67.2 L$

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