2C8H18(g)+25O2(g)-->16CO2(g)+18H2O(g)

Question

0.290mol of octane is allowed to react with 0.600mol of oxygen.The limiting reactant is Oxygen.
How many moles of water are produced in this reaction?

After the reaction, how much octane is left?

Answer 1

0.600 mole oxygen x (18 moles H2O/25 moles O2) = ?? moles H2O produced.

Use the same logic to convert 0.600 moles oxygen to moles octane, convert moles octane to grams octane, subtract from the grams you started with (of course convert 0.290 mols octane to grams) to determine grams remaining unreacted.

Answer 2

28.728