Liquid octane (CH3(CH2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 86g of octane is mixed with 147 g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits
To find the maximum mass of water that could be produced, we need to determine the limiting reactant first. This is the reactant that will be completely consumed and determines the amount of products formed.
1. Convert the given masses of octane and oxygen to moles using their molar masses:
- Mass of octane (C8H18) = 86 g
- Molar mass of octane = 8(12.01 g/mol) + 18(1.008 g/mol) = 114.22 g/mol
- Moles of octane = 86 g / 114.22 g/mol = 0.753 mol
- Mass of oxygen (O2) = 147 g
- Molar mass of oxygen = 2(16.00 g/mol) = 32.00 g/mol
- Moles of oxygen = 147 g / 32.00 g/mol = 4.59 mol
2. Write the balanced chemical equation to determine the stoichiometry of the reaction:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
According to the balanced equation, the stoichiometric ratio between octane and water is 2:18, or simplified, 1:9. Therefore, the moles of water produced should be 9 times the moles of octane.
3. Calculate the moles of water formed:
Moles of water = 0.753 mol octane × 9 = 6.777 mol
4. Convert moles of water to mass using the molar mass of water:
Molar mass of water = 2(1.008 g/mol) + 16.00 g/mol = 18.02 g/mol
Mass of water = 6.777 mol × 18.02 g/mol = 122.12 g
Rounded to 3 significant digits, the maximum mass of water that could be produced by the chemical reaction is 122 g.