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Question

11.2 g carbon reacts completely with 19.63 litre O2 at NTP. The cooled gases are passed through 2 litre of 2.5 N NaOH solution. Calculate concentration of remaining NaOH and Na2CO3 in solution. (CO does not react with NaOH under these conditions.)

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Solution

Let x moles of carbon be converted into CO and y moles of carbon be converted into CO2.
Cx+12O2x/2CO
Cy+O2yCO2
Total volume of oxygen used =x2×22.4+y×22.4
=19.63
11.2x+22.4y=19.63....(i)
x+y=11.212; i.e., 12x+12y=11.2....(ii)
Solving eqs. (i) and (ii), we get
x=0.11,y=0.82
Number of moles of CO2 formed =0.82
Number of milliequivalents of NaOH solution through which O2 is massed =N×V=2.5×2000=5000
Number of milliequivalents of CO2 passed =0.82×2×1000
=1640
2NaOH+CO2Na2CO3+H2O
Number of milliequivalents of Na2CO3=1640
NNa2CO3=16402000=0.82
Number of milliequivalent of remaining NaOH
=50001640=3360
Normality of remaining NaOH33602000=1.68.

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