Oleum is considered as a solution of `SO_(3)` in `H_(2)SO_(4)`, which is obtained by passing `SO_(3)` in solution of `H_(2)SO_(4)` When 100 g sample of oleum is diluted with desired mass of `H_(2)O` then the total mass of `H_(2)SO_(4)` obtained after dilution is known is known as % labelling in oleum.
For example, a oleum bottle labelled as '`109% H_(2)SO_(4)`' means the 109 g total mass of pure `H_(2)SO_(4)` will be formed when 100 g of oleum is diluted by 9 g of `H_(2)O` which combines with all the free `SO_(3)` present in oleum to form `H_(2)SO_(4)` as `SO_(3)+H_(2)O to H_(2)SO_(4)`
If excess water is added into a bottle sample labelled as "112%`H_(2)SO_(4)`" and is reacted with 5.3 g `NaCO_(3)` then find the volume of `CO_(2)` evolved at 1 atm pressure and 300 K temperature after the completion of the reaction :

Options

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

Answer

C

Explanation

`H_2SO_4+Na_2CO_3toNa_2SO_4+H_2O+CO_2`
moles of `CO_2` formed= mole of `Na_2CO_3` reacted (it is limiting reagent)`=5.3/106=0.05`
volume of `CO_2` formed at 1 atm pressure and 300 K=0.05x24.63=1.23 L