Answer to Question #21208 in Inorganic Chemistry for ekansh
(i) 0.6 ltr CO (ii) 0.8 ltr of CO2 (iii)0.6 ltr of CO2 and 0.8 ltr of CO (iv) None
CO2 + C = 2CO
According to equation, when 1 litre of CO2 reacts with 1 mol of coke, twolitres of CO are formed. Our mixture of gases has volume 1.4 L. Thus it is
composed from CO and non-reacted CO2.
If x litres of CO2 are reacted, 2x liters of CO are formed. Total volume ofmixture is the sum of CO2 and CO volumes:
V(mixture) = V(new volume CO2) + V(CO) = 1.4 L
V(new volume CO2) = V(initial volume CO2) - V(reacted CO2) = (1 - x) L
V(CO) = 2x
So, we have equation:
1.4 = 1 - x + 2x
The solution is x = 0.4 L
Thus, the mixture consist from 1-0.4=0.6 L of CO2 and 2*0.4=0.8 L of CO (iii)
The composition of product is 0.6 ltr of CO2 and 0.8 ltr of CO.
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If volume of mixture of CO2 and CO increases by 20% on passing over hot cake, then what is the percentage of CO2 in the original mixture?
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