The balanced equation for combustion of octane is shown: 2 C8H18 (I) + 25 O2 (g) —-> 16 CO2 (g) + 18 H2O (I) a. a gallon of octane has a mass of 2.661 x 10^3 grams. calculate the grams of carbon dioxide released by the combustion of one gallon of octane? b. calculate the number of moles of oxygen (O2 (g)) that would be needed to react with 2.4 moles of octane?
The Correct Answer and Explanation is:
Part a: Grams of Carbon Dioxide Released by the Combustion of One Gallon of Octane
To calculate the grams of carbon dioxide (CO2) released by the combustion of one gallon of octane, we need to follow these steps:
- Find the molar mass of octane (C8H18):
- Carbon (C) has a molar mass of 12.01 g/mol.
- Hydrogen (H) has a molar mass of 1.008 g/mol.
- For C8H18, the molar mass is calculated as: (8×12.01)+(18×1.008)=96.08+18.144=114.224 g/mol(8 \times 12.01) + (18 \times 1.008) = 96.08 + 18.144 = 114.224 \, \text{g/mol}(8×12.01)+(18×1.008)=96.08+18.144=114.224g/mol
- Calculate the moles of octane in 1 gallon:
- The mass of one gallon of octane is given as 2.661×103 g2.661 \times 10^3 \, \text{g}2.661×103g.
- The moles of octane are: 2.661×103 g114.224 g/mol≈23.3 mol\frac{2.661 \times 10^3 \, \text{g}}{114.224 \, \text{g/mol}} \approx 23.3 \, \text{mol}114.224g/mol2.661×103g≈23.3mol
- Use the balanced equation to find the mole ratio:
- From the balanced equation: 2 C8H18 (l)+25 O2 (g)→16 CO2 (g)+18 H2O (l)2 \, \text{C8H18 (l)} + 25 \, \text{O2 (g)} \rightarrow 16 \, \text{CO2 (g)} + 18 \, \text{H2O (l)}2C8H18 (l)+25O2 (g)→16CO2 (g)+18H2O (l)
- The mole ratio of octane to carbon dioxide is 2:162:162:16, meaning 2 moles of octane produce 16 moles of CO2.
- For 23.3 moles of octane, the moles of CO2 produced are: (162)×23.3≈186.4 mol of CO2\left( \frac{16}{2} \right) \times 23.3 \approx 186.4 \, \text{mol of CO2}(216)×23.3≈186.4mol of CO2
- Convert moles of CO2 to grams:
- The molar mass of CO2 is: (1×12.01)+(2×16.00)=44.01 g/mol(1 \times 12.01) + (2 \times 16.00) = 44.01 \, \text{g/mol}(1×12.01)+(2×16.00)=44.01g/mol
- The mass of CO2 produced is: 186.4 mol×44.01 g/mol≈8192.4 g186.4 \, \text{mol} \times 44.01 \, \text{g/mol} \approx 8192.4 \, \text{g}186.4mol×44.01g/mol≈8192.4g
- Therefore, the grams of CO2 released by the combustion of one gallon of octane is approximately 8192.4 grams.
Part b: Moles of Oxygen Needed to React with 2.4 Moles of Octane
To calculate the number of moles of oxygen (O2) required to react with 2.4 moles of octane, we can use the balanced equation:2 C8H18 (l)+25 O2 (g)→16 CO2 (g)+18 H2O (l)2 \, \text{C8H18 (l)} + 25 \, \text{O2 (g)} \rightarrow 16 \, \text{CO2 (g)} + 18 \, \text{H2O (l)}2C8H18 (l)+25O2 (g)→16CO2 (g)+18H2O (l)
From the equation, the mole ratio of octane to oxygen is 2:252:252:25. This means that for every 2 moles of octane, 25 moles of oxygen are required.
To find the moles of oxygen needed for 2.4 moles of octane:(252)×2.4≈30 mol of O2\left( \frac{25}{2} \right) \times 2.4 \approx 30 \, \text{mol of O2}(225)×2.4≈30mol of O2
Thus, 30 moles of O2 are needed to react with 2.4 moles of octane.
Summary:
- Grams of CO2 produced by one gallon of octane: 8192.4 grams.
- Moles of O2 needed to react with 2.4 moles of octane: 30 moles of O2.
