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Stolb23 [73]
4 years ago
10

When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limitin

g reactant and how much of the excess reactant is left over?
Chemistry
1 answer:
Nataliya [291]4 years ago
5 0
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
                                            
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