Chemistry - Mole Concept Question with Solution | TestHub

Question Explanation: Matching hydrocarbons with the relative amounts of CO₂ and H₂O produced upon complete combustion.

Concept: Stoichiometry of combustion.

Solution: The complete combustion of a hydrocarbon follows the general reaction:

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

We compare the coefficients of CO₂ ($x$) and H₂O ($y/2$) to determine the relative moles. The mass comparison depends on the molar masses.

(A) $C{H}_4+2O_2\to C{O}_2+2H_{2}O$

Moles of CO₂ = 1

Moles of H₂O = 2

Mass of CO₂ = 44 g

Mass of H₂O = $18\times 2=36$ g

(B) $C_2{H}_4+3O_2\to 2C{O}_2+2H_{2}O$

Moles of CO₂ = 2

Moles of H₂O = 2

Mass of CO₂ = $44\times 2=88$ g

Mass of H₂O = $2\times 18=36$ g

(C) $C_2{H}_2+\frac{5}{2}{O}_2\to 2C{O}_2+H_{2}O$

Moles of CO₂ = 2

Moles of H₂O = 1

Mass of CO₂ = $2\times 44=88$ g

Mass of H₂O = $1\times 18=18$ g

(D) $C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$

Moles of CO₂ = 3

Moles of H₂O = 4

Mass of CO₂ = $3\times 44=132$ g

Mass of H₂O = $4\times 18=72$ g

Final Answer: (A)-(P,S), (B)-(Q, S), (C)-(R,S), (D)-(P,S)