When `12.0g` of carbon (graphite)reacted with oxygen to form `CO` and `CO_(2)` at `25^(@)C` and constant pressure, `252 kJ` of heat was released and no carbon remained. If `Delta H_(f)^(0)(CO,g)= -110.5 kJ mol^(-1)` and `Delta H_(f)^(0)(CO_(2),g)= -393.5 kJ mol^(-1)`,calculate the mass of oxygen consumed.

Answer

`24 g`

Explanation

We have
Amount of carbon `=(12.0g)/(12.0g mol^(-1))=1 mol`
The equation to be considered are
`C("graphite")=(1)/(2)O_(2)(g)rarrCO(g)DeltaH^(0)= -110.5 kJ mol^(-1)`
`C("graphite")+O_(2)(g)rarrCO_(2)(g)DeltaH^(0)= -393.5 kJ mol^(-1)`
Let the amount `x` of carbon be converted into `CO` and the remaining `(i.e. 1.0 mol-x)` into `CO_(2)`. we will have `[(x(-110.5)+(1.0 mol-x)(-393.5)]kJ mol^(-1) = -313.8 kJ`
Which gives `x=(252+393.5)/(393.5-110.5)mol=0.5 mol`
Amount of oxygen needed `=[(0.5)/(2)+(1.0-0.5)]mol=0.75 mol`
Mass of oxygen needed `=(0.75 mol)(32 g mol^(-1))=24 g`