Question

# Oleum is considered as a solution of $$SO_3$$ in $$H_2SO_4$$. Which is obtained by passing $$SO_3$$ in solution of $$H_2SO_4$$.When 100 g sample of oleum is diluted with desired weight of $$H_2O$$ then the total mass of $$H_2SO_4$$.obtained after dilution is known as labelling of oleum. For example , a oleum is diluted by 9 g of $$H_2O$$ which combined with all the free $$SO_3$$ to form $$H_2SO_4$$ as $$SO_3 +H_2O \rightarrow H_2SO_4$$.In excess water is added into a 100 g bottle sample labelled as 112 %  $$H_2SO_4$$ and is reacted with 5.3 g $$Na_2CO_3$$, then find the volume of $$CO_2$$ evolved at 1 atm pressure and 300 K temperature after the completion of the reaction. [R=$$0.0821 L \, atm \, mol^{-1} \, k^{-1}]$$.$$H_2SO_4 +Na_2CO_3 \rightarrow Na_2SO_4 +H_2O+CO_2$$

A
2.46 L
B
24.6 L
C
1.23 L
D
12.3 L

Solution

## The correct option is C 1.23 L$$H_2SO_4 +Na_2CO_3 \rightarrow Na_2SO_4 +H_2O +CO_2$$Moles of $$CO_2$$ formed =moles of $$Na_2CO_3$$ reacted ( is is limiting reagent )= $$\dfrac {5.3}{106}=0.05$$Volumes of $$CO_2$$ formed at  1 atm  pressure and $$300 K =0.05 \times 24.63=1.23 L$$Chemistry

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