Given the equation 2C8H18 + 25O2 = 16 CO2 + 18H2O: The combustion of octane produces carbon dioxide and water. If you start with 10.0 moles of octane, what mass of water (in grams) would you expect to produce?
To determine the mass of water produced in the combustion of octane, we first need to find the molar ratio between octane (C8H18) and water (H2O) in the balanced chemical equation.
The balanced equation is:
2C8H18 + 25O2 → 16CO2 + 18H2O
From the equation, we can see that every 2 moles of octane (C8H18) will produce 18 moles of water (H2O). This means the mole ratio between octane and water is 2:18 or 1:9.
Now, let's calculate the number of moles of water produced from 10.0 moles of octane.
Moles of water = moles of octane * (moles of water / moles of octane)
= 10.0 moles * (18 moles / 2 moles)
= 90.0 moles
So, 10.0 moles of octane would produce 90.0 moles of water.
Next, we need to convert the moles of water to grams using the molar mass of water (H2O), which is approximately 18.015 g/mol.
Mass of water = moles of water * molar mass of water
= 90.0 moles * 18.015 g/mol
≈ 1621.35 grams
Therefore, if you start with 10.0 moles of octane, you would expect to produce approximately 1621.35 grams of water.