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Limiting Reagent or Reactant

If excess wat...

Question

If excess water is added into a $100 g$ bottle sample labelled as $" 112 % H_{2} S O_{4} "$ and is reacted with $5.3 g$ $N a_{2} C O_{3}$ , then find the volume of $C O_{2}$ evolved at STP after the completion of the reaction : [ $R = 0.0821 L a t m m o l$ $^{- 1} K^{- 1}$ ]
$H_{2} S O_{4} + N a_{2} C O_{3} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$

2.46 L

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24.6 L

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1.12 L

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11.2 L

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Solution

The correct option is C 1.12 L
Balanced chemical reaction :
$H_{2} S O_{4} + N a_{2} C O_{3} \to N a_{2} S O_{4} + H_{2} O + C O_{2}$

Moles = $\frac{G i v e n m a s s}{M o l a r m a s s}$
Moles of $N a_{2} C O_{3}$ = $\frac{5.3}{106} = 0.05 m o l$
Moles of $H_{2} S O_{4}$ = $\frac{112}{98} = 1.14 m o l$
For limiting reagent = $\frac{M o l e s}{S t o c h i o m e t r i c c o e f f i c e n t}$
For $N a_{2} C O_{3}$ = $\frac{0.05}{1} = 0.05$
For $H_{2} S O_{4}$ = $\frac{112}{98} = 1.14$ .
The ratio for $N a_{2} C O_{3}$ is less hence it is a limiting reagent.
According to stoichiometric coefficient :
Moles of $C O_{2}$ formed = Moles of $N a_{2} C O_{3}$ reacted = 0.05 mol.
Volume of $C O_{2}$ evolved at STP conditions = $0.05 \times 22.4 = 1.12 L$

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Similar questions

Q. If excess water is added into a

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and is reacted with

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Limiting Reagent or Reactant

Standard XIII Chemistry

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If excess water is added into a 100 g bottle sample labelled as "112% H2SO4" and is reacted with 5.3 g Na2CO3, then find the volume of CO2 evolved at STP after the completion of the reaction : [R=0.0821 L atm mol−1 K−1] H2SO4+Na2CO3→Na2SO4+H2O+CO2