Question

The combustion of ethanol (C2H5OH) takes place by the following reaction equation.

C2H5OH (l) + 3 O2 (g) → 2 CO2 (g) + 3 H2O (g)

What is the volume of CO2 gas produced by the combustion of excess ethanol by 23.3 grams of O2 gas at 25°C and 1.25 atm?

Answer 1

`\boxed{\text{9.50 L}}`

Step-by-step 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

`\text{Moles of O}_(2) =\text{23.3 g O}_(2) * \frac{\text{1 mol O}_(2)}{\text{32.00 g O}_(2)} =\text{0.7821 mol O}_(2)`

(b) Moles of CO₂

The molar ratio is 2 mol CO₂ = 3 mol O₂

`\text{Moles of CO$_(2)$}= \text{0.7821 mol O}_(2) * \frac{\text{2 mol CO$_(2)$}}{ \text{3 mol O}_(2)} = \text{0.4854 mol CO$_(2)$}`

(c) Volume of CO₂

We can use the Ideal Gas Law to calculate the volume.

T = 25 °C = 298.15 K

`\begin{array}{rcl}pV & = & nRT\\1.25V & = & 0.4854 * 0.08206 * 298.15\\1.25V & = & 11.88\\V & = & \textbf{9.50 L}\\\end{array}\\\text{The volume of CO$_(2)$ produced is $\boxed{\textbf{9.50 L}}$}`