Question

How many gram of CO2 is produced by the combustion of 44 gm of
propane.​

Answer 1

Answer:

       132gm

Explanation:

propane = C₃H₈

Molecular mass of propane = (3×12) + (8×1) = 36 + 8 = 44gm

CₓHₙ + (x + n/4)O₂ ⇒ xCO₂ + n/2H₂O   (general formula of combustion)

C₃H₈ + (3 + 8/4)O₂ ⇒ 3CO₂ + 8/2H₂O

C₃H₈ + (3 + 2)O₂ ⇒ 3CO₂ + 4H₂O

C₃H₈ + 5O₂ ⇒ 3CO₂ + 4H₂O

44gm C₃H₈ ≡ 3mol CO₂

No of moles = given mass (w) / Molecular mass

3 = w / 44

w = 44 × 3

w = 132gm

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