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The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a. Calcluate the volume of carbon dioixde ( at s.T.P.) that would be produced by the combustion of 27.5 g of C3H8 burns in the presence of 45.0 L of O2.

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Answer:

26.9 L is the volume of COâ‚‚, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of Oâ‚‚ to determine mass and then, the moles.

Oâ‚‚ density = Oâ‚‚ mass / Oâ‚‚ volume

Oâ‚‚ density . Oâ‚‚ volume = Oâ‚‚ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of Oâ‚‚

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of Oâ‚‚ produce 3 moles of COâ‚‚

Therefore, 2.009 moles of Oâ‚‚ must produce (2.009 .3) /5 = 1.21 moles of COâ‚‚. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

1 atm . V = 1.21 moles . 0.082 . 273K

V = (1.21 moles . 0.082 . 273K) / 1atm = 26.9 L

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