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11.2 g of carbon reacts with 21.1 litres...

`11.2 g` of carbon reacts with 21.1 litres of oxygen at `18^(@)C` and `750 mm` of `Hg`. The cooled gases are passed through 2 litre of `2.5 N NaOH` . Assume that `CO` does not react with `NaOH`:
a. What is the mole fraction of `CO` in the gases?

Text Solution

Verified by Experts

Mole of `C = (11.2)/(12) = 0.933 mol`
`Pv = nRT`
`(750)/(760) xx 21.2 = n xx 0.82 xx 291`
`n = (750 xx 21.2)/(760 xx 0.082 xx 291) = 0.876 mol of O_(2)`
a. `C + O_(2) rarr CO_(2)`
`x` `x` `x`
b. `C + (1)/(2) O_(2) rarr CO`
`(0.933 - x) (0.933 - x)/(2) (0.933 - x)`
Total moles of `O_(2) = x + (0.933 - x)/(2) = 0.876`
`:. x = 0.821 mol`
`CO_(2) = 0.821 mol`
`CO = 0.933 - 0.821= 0.112 mol`
a. `:. x_(CO) = (n_(CO))/(n_(CO) + n_(CO_(2))) = (0.122)/(0.93) = 0.12`
`underset({:(1 mol),(0.821 mol):})(CO_(2)) + underset({:(2 mol),(2 xx 0.821 mol):})(2NaOH) rarr underset(1 mol)(Na_(2)CO_(3)) + H_(2)O`
Total moles of `NaOH = 2 xx 821`
Moles of `NaOH` left `= 5 - 2 xx 0.821`
`= 3.558 mol`
Concentration of `NaOH = ("Moles")/("Volume") = (3.385)/(2)`
`= 1.679 mol L^(-1)`
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