The balanced reaction equation for the reaction between CH₃OH and O₂ is
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles 12 24
Reacted moles 12 18
Final moles - 6 12 24
The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
= 12 mol x 3 / 2
= 18 mol
All of CH₃OH moles react with O₂.
Hence, the limiting agent is CH₃OH.
Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol
If the temperature of a liquid-vapor system at equilibrium increases, it will shift towards the vapor phase, assuming that the pressure remains equal. The concentration of vapor will also increase relative to the concentration of liquid in the system. Thus, the new equilibrium condition will have more vapor than liquid.
Hydrogen mass is 1, the atom with mass number 7 is Lithium
The answer will be 3 moles
