Questions: Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 33. g of octane is mixed with 168. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 33. g of octane is mixed with 168. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Transcript text: Liquid octane $\left(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{3}\right)$ will react with gaseous oxygen $\left(\mathrm{O}_{2}\right)$ to produce gaseous carbon dioxide $\left(\mathrm{CO}_{2}\right)$ and gaseous water $\left(\mathrm{H}_{2} \mathrm{O}\right)$. Suppose 33. g of octane is mixed with 168. g of oxygen. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction. Be sure your answer has the correct number of significant digits.

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the combustion of octane is:

\[ 2 \mathrm{C}_8\mathrm{H}_{18} + 25 \mathrm{O}_2 \rightarrow 16 \mathrm{CO}_2 + 18 \mathrm{H}_2\mathrm{O} \]

Step 2: Calculate Molar Masses

Calculate the molar masses of octane ($\mathrm{C}_8\mathrm{H}_{18}$) and oxygen ($\mathrm{O}_2$):

Molar mass of $\mathrm{C}_8\mathrm{H}_{18}$: \[ (8 \times 12.01) + (18 \times 1.008) = 114.22 \, \text{g/mol} \]
Molar mass of $\mathrm{O}_2$: \[ 2 \times 16.00 = 32.00 \, \text{g/mol} \]

Step 3: Determine Moles of Reactants

Calculate the moles of octane and oxygen:

Moles of $\mathrm{C}_8\mathrm{H}_{18}$: \[ \frac{33.0 \, \text{g}}{114.22 \, \text{g/mol}} = 0.2888 \, \text{mol} \]
Moles of $\mathrm{O}_2$: \[ \frac{168.0 \, \text{g}}{32.00 \, \text{g/mol}} = 5.250 \, \text{mol} \]

Step 4: Identify the Limiting Reactant

Using the stoichiometry from the balanced equation, determine the limiting reactant:

According to the equation, 2 moles of $\mathrm{C}_8\mathrm{H}_{18}$ require 25 moles of $\mathrm{O}_2$. Therefore, 0.2888 moles of $\mathrm{C}_8\mathrm{H}_{18}$ would require: \[ 0.2888 \times \frac{25}{2} = 3.610 \, \text{mol of } \mathrm{O}_2 \]

Since 3.610 moles of $\mathrm{O}_2$ are required and we have 5.250 moles available, $\mathrm{C}_8\mathrm{H}_{18}$ is the limiting reactant.

Step 5: Calculate Maximum Mass of $\mathrm{CO}_2$

Using the stoichiometry of the balanced equation, calculate the moles of $\mathrm{CO}_2$ produced from the limiting reactant:

From 2 moles of $\mathrm{C}_8\mathrm{H}_{18}$, 16 moles of $\mathrm{CO}_2$ are produced. Therefore, 0.2888 moles of $\mathrm{C}_8\mathrm{H}_{18}$ will produce: \[ 0.2888 \times \frac{16}{2} = 2.310 \, \text{mol of } \mathrm{CO}_2 \]

Calculate the mass of $\mathrm{CO}_2$:

Molar mass of $\mathrm{CO}_2$: \[ (12.01) + (2 \times 16.00) = 44.01 \, \text{g/mol} \]
Mass of $\mathrm{CO}_2$: \[ 2.310 \, \text{mol} \times 44.01 \, \text{g/mol} = 101.7 \, \text{g} \]