Liquid octane (CH3(Ch2)6CH3) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO2) and gaseous water (H2O). Suppose 75. g of octane is mixed with 153. g of oxygen. Calculate the maximum mass of water that could be produced by the chemical reaction. Round your answer to 3 significant digits.
The balanced chemical equation for the reaction between octane and oxygen is:
2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O
To determine the maximum mass of water that could be produced, we need to calculate the limiting reactant. This can be done by comparing the moles of each reactant.
The molar mass of octane (C8H18) is 114.23 g/mol.
The molar mass of oxygen (O2) is 32.00 g/mol.
First, we calculate the moles of octane:
moles of octane = 75 g / 114.23 g/mol = 0.656 mol
Next, we calculate the moles of oxygen:
moles of oxygen = 153 g / 32.00 g/mol = 4.781 mol
According to the balanced equation, 2 moles of octane react with 25 moles of oxygen to produce 18 moles of water.
Using the mole ratio, we can determine the maximum moles of water that could be produced:
moles of water = (0.656 mol octane) * (18 mol H2O / 2 mol octane) = 5.912 mol H2O
Finally, we calculate the mass of water:
mass of water = (5.912 mol H2O) * (18.02 g/mol) = 106.4 g
Therefore, the maximum mass of water that could be produced is 106.4 g. Rounded to three significant digits, the answer is 106 g.