Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?

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Montes

Chemistry

28 August, 02:04

Gas stoichiometry: what volume of oxygen at 25 degrees celsius and 1.04 atm is needed for the complete combustion of 5.53 grams of propane?

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Keenan · Answer 1

Answer is: volume of oxygen is 14.7 liters.

Balanced chemical reaction: C₃H₈ + 5O₂ → 3CO₂ + 4H₂O.

m (C₃H₈-propane) = 5.53 g.

n (C₃H₈) = m (C₃H₈) : M (C₃H₈).

n (C₃H₈) = 5.53 g : 44.1 g/mol.

n (C₃H₈) = 0.125 mol.

From chemical reaction: n (C₃H₈) : n (O₂) = 1 : 5.

n (O₂) = 0.625 mol.

T = 25° = 298.15K.

p = 1.04 atm.

R = 0.08206 L·atm/mol·K.

Ideal gas law: p·V = n·R·T.

V (O₂) = n·R·T / p.

V (O₂) = 0.625 mol · 0.08206 L·atm/mol·K · 298.15 K / 1.04 atm.

V (O₂) = 14.7 L.