11.2 g carbon reacts completely with 19.63 litre `O_(2)` at NTP. The cooled gases are pased through 2 litre of 2.5 N NaOH solution. Calculate concentration of remaining NaOH and `Na_(2)CO_(3)` in solution. (CO does not react with NaOH under these conditions.)

Explanation

Let x moles of carbon be converted into CO and y moles of carbon be converted into `CO_(2)`.
`underset(x)(C )+(1)/(2)underset(x//2)(O_(2)) to CO`
`underset(y)C+underset(y)(O_(2)) to CO_(2)`
Total volume of oxygen used `=(x)/(2)xx22.4+yxx22.4`
=19.63
11.2x+22.4y=19.63
`x+y=(11.2)/(12)`, i.e., 12x+12y=11.2
Solving eqs. (i) and (ii), we get
x=0.11, y=0.82
Number of moles of `CO_(2)` formed =0.82
Number of milliequivalents of NaOH solution through which `CO_(2)` is massed `=NxxV=2.5xx2000=5000`
Number of milliequivalents of `CO_(2)` passed`=0.82xx2xx1000`
=1640
`2NaOH+CO_(2) to Na_(2)CO_(3)+H_(2)O`
Number of milliequivalents of `Na_(2)CO_(3)=1640`
`N_(Na_(2)CO_(3))=(1640)/(2000)=0.82`
Number of milliequivalents of remaining NaOH
=5000-1640=3360
Normality of remaining `NaOH=(3360)/(2000)=1.68`