Answer:
26.9 L is the volume of CO₂, 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 O₂ to determine mass and then, the moles.
O₂ density = O₂ mass / O₂ volume
O₂ density . O₂ volume = O₂ 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 O₂
Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles
Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂
Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. 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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