How many grams of water (h2o) are produced if i burn 4 moles of octane (c8h18) in excess oxygen? 2c8h18 + 25o2  16co2 + 18h2o

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

i believe it is c.) 1575 j.

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