Questions: Given the equation C2H6(g)+O2(g) → CO2(g)+H2O(g) (not balanced). determine the number of liters of O2 consumed at STP when 76.7 grams of C2H6 is burned. 3.42 L 181 x 10^5 1 57.1 L 224 L

Given the equation C2H6(g)+O2(g) → CO2(g)+H2O(g) (not balanced). determine the number of liters of O2 consumed at STP when 76.7 grams of C2H6 is burned.
3.42 L
181 x 10^5 1
57.1 L
224 L

Transcript text: Given the equation $\mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (not balanced). determine the number of fiters of $\mathrm{O}_{2}$ consumed at STP when 76.7 grams of $\mathrm{C}_{2} \mathrm{H}_{6}$ is burned. 3.42 L $181 \times 10^{5} 1$ 57.1 L 224 L

Solution

Solution Steps

Step 1: Balance the Chemical Equation

First, we need to balance the chemical equation for the combustion of ethane ($\mathrm{C}_2\mathrm{H}_6$):

\[ \mathrm{C}_2\mathrm{H}_6(g) + \mathrm{O}_2(g) \rightarrow \mathrm{CO}_2(g) + \mathrm{H}_2\mathrm{O}(g) \]

Balancing the equation, we get:

\[ 2\mathrm{C}_2\mathrm{H}_6(g) + 7\mathrm{O}_2(g) \rightarrow 4\mathrm{CO}_2(g) + 6\mathrm{H}_2\mathrm{O}(g) \]

Step 2: Calculate Moles of $\mathrm{C}_2\mathrm{H}_6$

Next, calculate the moles of $\mathrm{C}_2\mathrm{H}_6$ using its molar mass. The molar mass of $\mathrm{C}_2\mathrm{H}_6$ is approximately $30.07 \, \text{g/mol}$.

\[ \text{Moles of } \mathrm{C}_2\mathrm{H}_6 = \frac{76.7 \, \text{g}}{30.07 \, \text{g/mol}} \approx 2.550 \, \text{mol} \]

Step 3: Determine Moles of $\mathrm{O}_2$ Required

From the balanced equation, 2 moles of $\mathrm{C}_2\mathrm{H}_6$ require 7 moles of $\mathrm{O}_2$. Therefore, 1 mole of $\mathrm{C}_2\mathrm{H}_6$ requires $3.5$ moles of $\mathrm{O}_2$.

\[ \text{Moles of } \mathrm{O}_2 = 2.550 \, \text{mol} \times 3.5 = 8.925 \, \text{mol} \]

Step 4: Convert Moles of $\mathrm{O}_2$ to Liters at STP

At standard temperature and pressure (STP), 1 mole of any gas occupies $22.414 \, \text{L}$.

\[ \text{Liters of } \mathrm{O}_2 = 8.925 \, \text{mol} \times 22.414 \, \text{L/mol} \approx 200.0 \, \text{L} \]