Consider the following reactions:
`C(s)+O_(2)(g)toCO_(2)(g)+x " kJ"`
`CO(g)+(1)/(2)O_(2)(g)toCO_(2)(g)+y" kJ"`
The heat formation of CO(g) is :

Explanation

(i) Weight of coal = 100 kg
Weight of carbon in coal = `(100 xx80)/(100) = 80 kg`
Weight of carbon converted to `CO_(2) = (80 xx 60)/(100) = 48 kg`
Weight of carbon converted to `CO +(80 xx 40)/(100) = 32` kg
Now according to reaction,
`C(s) +O_(2)(g) to CO_(2)(g) + 394 kJ`
12 g of carbon on combustion to `CO_(2)` give heat = 394 kJ
`48 xx 10^(3)` g of carbon on combustion to `CO_(2)` give heat = `(394 xx 48 xx 10^(3))/(12) = 1576 xx 10^(3)`kJ Similarly,
`C(S) + 1/2O_(2)(g) to CO(g) + 111 kJ`
`12 g` of carbon on combustion to CO give heat = 111 kJ
`32 xx 10^(3)`g of carbon on combustion to CO give heat = `(111xx32xx10^(3))/(12) = 296 xx 10^(3)`kJ
`therefore` Total heat evolved = `1576 xx 10^(3) + 296 xx 10^(3)`
`= 1872 xx 10^(3)`kJ .
(ii)When more efficient oven is used, only `CO_(2)` will be formed from 80 kg of carbon : 12 g of carbon on combustion to `CO_(2)` give heat = 394 kJ
`80 xx 10^(3)` g of carbon on combustion to `CO_(2)` give heat =` (394 xx 80 xx 10^(3))/(12) = 2626.7 xx 10^(3)kJ`
(iii) Percentage loss in heating value of inefficient oven may be calculated as :
Percentage loss `= (2626.7 xx 10^(3) - 1872 xx 10^(3))/(2626.7 xx 10^(3)) xx 100 = 28.7%`