Answer:
[tex]\boxed{\text{9.50 L}}[/tex]
Explanation:
We will need a balanced chemical equation with masses and molar masses, volumes, and concentrations, so, let's gather all the information in one place.
M_r: 32.00
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(g)
m/g: 23.3
(a) Moles of O₂
[tex]\text{Moles of O}_{2} =\text{23.3 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} =\text{0.7821 mol O}_{2}[/tex]
(b) Moles of CO₂
The molar ratio is 2 mol CO₂ = 3 mol O₂
[tex]\text{Moles of CO$_{2}$}= \text{0.7821 mol O}_{2} \times \dfrac{\text{2 mol CO$_{2}$}}{ \text{3 mol O}_{2}} = \text{0.4854 mol CO$_{2}$}[/tex]
(c) Volume of CO₂
We can use the Ideal Gas Law to calculate the volume.
T = 25 °C = 298.15 K
[tex]\begin{array}{rcl}pV & = & nRT\\1.25V & = & 0.4854 \times 0.08206 \times 298.15\\1.25V & = & 11.88\\V & = & \textbf{9.50 L}\\\end{array}\\\text{The volume of CO$_{2}$ produced is $\boxed{\textbf{9.50 L}}$}[/tex]
