How many grams of water (H2O) are produced if I burn 4 moles of octane (C8H18) in excess oxygen? 2C8H18 + 25O2  16CO2 + 18H2O

Answer: 25.7 mol

From the balanced reaction, 2 mol of Octane produces 18 mol of Water.

Find the moles of Octane you are reacting.

moles of Octane = mass of Octane / molar mass = 325g / 114.23 g/mol = 2.85 mol.

Since Oxygen is in excess, limiting reagent is Octane.

2 mol Octane produces 18 mol Water.

1mol Octane produces 9 mol Water.

Therefore, 2.85 mol Octane will produce Water = 2.85 mol Octane x 9 mol Water / 1mol Octane

= 25.7 mol