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 with final answer
To calculate the maximum mass of water that could be produced, we need to determine the limiting reactant first.
1. Calculate the number of moles for each substance.
- Moles of octane: 86g / molar mass of octane.
The molar mass of octane (C8H18) can be calculated as follows:
(12.01 g/mol * 8) + (1.01 g/mol * 18) = 114.23 g/mol.
Moles of octane = 86 g / 114.23 g/mol = 0.752 mol.
- Moles of oxygen: 147g / molar mass of oxygen.
The molar mass of oxygen (O2) is 32.00 g/mol.
Moles of oxygen = 147 g / 32.00 g/mol = 4.59 mol.
2. Write the balanced equation:
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
3. Determine the limiting reactant:
The stoichiometric ratio between octane and oxygen is 2:25. Thus, 2 moles of octane require 25 moles of oxygen.
The number of moles of oxygen available (4.59 mol) is higher than the number of moles required for 2 moles of octane. Therefore, oxygen is in excess, and octane is the limiting reactant.
4. Calculate the maximum mass of water:
From the stoichiometry of the balanced equation, we know that 2 moles of octane produce 18 moles of water.
Moles of water = 0.752 mol (moles of octane) * (18 moles of water / 2 moles of octane) = 6.768 mol.
Finally, to determine the maximum mass of water:
Mass of water = moles of water * molar mass of water.
The molar mass of water (H2O) is (1.01 g/mol * 2) + 16.00 g/mol = 18.02 g/mol.
Mass of water = 6.768 mol * 18.02 g/mol = 122.04 g.
Therefore, the maximum mass of water that could be produced by the chemical reaction is 122.04 grams, rounded to 3 significant digits.