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.
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
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