Answer:
12 moles of H₂O are formed in this combustion.
Explanation:
First of all, think the reaction:
2CH₃OH (l) + 3O₂ (g) → 2CO₂ (g) + 4H₂O (g)
Ratio in the reactants is 2:3, so 2 mol of methanol need 3 mol of oxygen to react. Then 8 mol of CH₃OH, will need (8.3)/2 = 12 moles of O₂
We have 9 moles of O₂, so this is the limiting reactant.
3 mol of oxygen produce 4 mol of water
Then, 9 mol of oxygen will produce ( 9 .4)/3 = 12 moles
