The equation for the complete combustion ch₄ is: ch₄ + 2o₂ → co₂ + 2h₂o if 3.70 moles of methane reacted with 5.40 moles of oxygen, what is the limiting reactant in the formation of water?

Explanation:

                           CH₄          +            2O₂ → CO₂ + 2H₂O

from equation    1 mol                      2 mol

given                   3.70 mol               5.40 mol

1 mol CH₄  needs 2 mol O₂

3.70 mol CH₄   need   x mol  O₂

x=3.70*2/1=7.40 mol O₂

So, we see that 3.70 mol CH₄ need 7.40 mol O₂, but only 5.40 mol O₂ given (it is not enough).

So, O₂ is the limiting reactant.

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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