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Limiting Reagent or Reactant

6 moles of Na...

Question

6 moles of $N a_{2} C O_{3}$ is reacted with 10 moles of $H C l$ . Find the volume of $C O_{2}$ gas produced at STP.
The reaction is:
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + C O_{2} + H_{2} O$

$134.4 L$

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112 L

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22.4 L

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44.8 L

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Solution

The correct option is B $112 L$
Given reaction is:
$N a_{2} C O_{3} + 2 H C l \to 2 N a C l + C O_{2} + H_{2} O$
Divide the given moles of reactant by its the respective stoichiometric coefficient. So, ratio of
$N a_{2} C O_{3} =$ $\frac{6}{1} = 6$ , $H C l = \frac{10}{2} = 5$ (division is minimum)
Therefore, $H C l$ is the limiting reagent.
So, $mol of \frac{H C l}{2} = mol of C O_{2} produced$
Mole of $C O_{2}$ produced = 5 moles
Volume of $C O_{2}$ produced at S.T.P. $= 5 \times 22.4 = 112 L$

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