2C8H18(g)+25O2(g)+16CO2(g)+18H2O(g)
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?
No arrow, no products or reactants.
To determine the moles of water produced and the remaining moles of octane, we need to use the stoichiometry of the balanced chemical equation.
The balanced equation for the reaction is:
2C8H18(g) + 25O2(g) → 16CO2(g) + 18H2O(g)
Given:
Moles of octane (C8H18) = 0.290 mol
Moles of oxygen (O2) = 0.600 mol
Based on the balanced equation, we can see that the stoichiometric ratio between oxygen and water is 25:18. Since the limiting reactant is oxygen, it will be completely consumed, and the moles of water formed will be determined by the stoichiometry.
To find the moles of water produced:
Moles of oxygen used = Moles of oxygen given = 0.600 mol
Moles of water produced = (18/25) * Moles of oxygen used = (18/25) * 0.600 mol = 0.432 mol
Therefore, 0.432 moles of water are produced in this reaction.
To determine how much octane is left, we need to determine the moles of octane consumed. Based on the balanced equation, we can see that the stoichiometric ratio between octane and oxygen is 2:25. Therefore, the moles of octane consumed will be determined by the stoichiometry.
Moles of oxygen consumed = Moles of oxygen given = 0.600 mol
Moles of octane consumed = (2/25) * Moles of oxygen consumed = (2/25) * 0.600 mol = 0.048 mol
To find the moles of octane remaining:
Moles of octane remaining = Moles of octane given - Moles of octane consumed
Moles of octane remaining = 0.290 mol - 0.048 mol = 0.242 mol
Therefore, after the reaction, 0.242 moles of octane are left.