JEE Main 2022 — Some Basic Concepts Of Chemistry Question with Solution

$$C_x{H}_y+\left(x+\frac{y}{4}\right)O_2\to xC{O}_2+\frac{y}{2}{H}_{2}O$$

From the reaction,

Produced CO₂ = x mol

and produced H₂O = $$\frac{y}{2}$$ mol

Given produced CO₂ = 330 g

$$\therefore$$ moles of CO₂ $$=\frac{330}{44}=\frac{30}{4}=x$$

Also given produced H₂O = 270 gm

$$\therefore$$ Moles of H₂O $$=\frac{270}{18}=15=\frac{y}{2}$$.

$$\Rightarrow$$ y = 30

$$\therefore$$ $$x:y=\frac{30}{4}:30=1:4$$

Formula of the compound = $$(C{H}_4{)}_n$$

$$\therefore$$ Weight of C in $$(C{H}_4{)}_n$$ = 12 n

Weight of H in $$(C{H}_4{)}_n$$ = 4 n

$$\therefore$$ Weight ratio of C and H

= 12 n : 4 n

= 3 : 1

$$\therefore$$ % of C = $$\frac{3}{4}$$ $$\times$$ 100 = 75

and % of H = $$\frac{1}{4}$$ $$\times$$ 100 = 25