Answer:
[tex]\large \boxed{\text{27.0 L CO}_{2}}[/tex]
Explanation:
We are given the amounts of two reactants and asked to determine the amount of product.
This is a limiting reactant problem.
1. Assemble the information
We will need a balanced chemical equation with molar masses and volumes, so, let's gather all the information in one place.
MV/L: 22.71 22.71
Mᵣ: 44.10
C₃H₈ + 5O₂ ⟶ 3CO₂ + 4H₂O
m/g: 27.5
V/L: 45.0
2. Calculate the moles of each reactant
(a) Propane
[tex]\text{Moles of C$_{3}$H$_{8}$} = \text{27.5 g C$_{3}$H$_{8}$} \times \dfrac{\text{1 mol C$_{3}$H$_{8}$}}{\text{44.10 g C$_{3}$H$_{8}$}} = \text{0.6236 mol C$_{3}$H$_{8}$}[/tex]
(b) Oxygen
The volume of 1 mol of a gas at STP (0 °C and 1 bar) is 22.71 L
[tex]\text{Moles of O$_{2}$} = \text{45.0 L O}_{2} \times \dfrac{\text{1 mol O$_{2}$}}{\text{22.71 L O$_{2}$}} = \text{1.982 mol O$_{2}$}[/tex]
3. Calculate the moles of CO₂ from each reactant
[tex]\textbf{From C$_{3}$H$_{8}$:}\\\text{Moles of CO$_{2}$} = \text{0.6236 mol C$_{3}$H$_{8}$} \times \dfrac{\text{3 mol CO$_{2}$}}{\text{1 mol C$_{3}$H$_{8}$}} = \text{1.871 mol CO}_{2}\\\\\textbf{From O$_{2}$:}\\\text{Moles of CO$_{2}$} =\text{1.982 mol O$_{2}$} \times \dfrac{\text{3 mol CO$_{2}$}}{\text{5 mol O$_{2}$}} = \text{1.189 mol CO$_{2}$}\\\\\textbf{Oxygen} \text{ is the limiting reactant because it gives fewer moles of CO$_{2}$.}[/tex]
4. Calculate the volume of CO₂
[tex]\text{ Volume of CO$_{2}$} = \text{1.189 mol CO$_{2}$} \times \dfrac{\text{22.71 L CO$_{2}$}}{\text{1 mol CO$_{2}$}} = \textbf{27.0 L CO}_\mathbf{{2}}\\\\\text{The reaction produces $\large \boxed{\textbf{27.0 L CO}_\mathbf{{2}}}$}[/tex]
