When can we use the molar ratio?

The mole ratios that exist between the chemical species that take part in a reaction are simply the stoichiometric coefficients needed to balance the chemical equation that describes said reaction.

Mind you, it doesn't matter if the reaction goes to completion or if you're dealing with an equilibrium reaction, but it does matter if the chemical equation is balanced or not.

More specifically, you can only use mole ratios when the chemical equation is balanced.

Let's take, for example, the reaction

#"CH"_ (4(g)) + color(blue)(2)"O"_ (2(g)) -> "CO"_ (2(g)) + color(purple)(2)"H"_ 2"O"_ ((l))#

Notice that the reaction consumes #color(blue)(2)# moles of oxygen gas for every #1# mole of methane that undergoes combustion and produces #1# mole of carbon dioxide and #color(purple)(2)# moles of water.

These mole ratios tell you that regardless of how many moles of one reactant you have, there will always be a #1:color(blue)(2)# mole ratio between them.

Similarly, the reaction will always produce twice as many moles of water than of carbon dioxide because the two reactants are in a #color(purple)(2):1# mole ratio.

So, if #7# moles of methane react, you will have

#7 color(red)(cancel(color(black)("moles CH"_4))) * (color(blue)(2)color(white)(.)"moles O"_2)/(1color(red)(cancel(color(black)("mole CH"_4)))) = "14 moles O"_2 -># needed

#7 color(red)(cancel(color(black)("moles CH"_4))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole CH"_4)))) = "7 moles CO"_2 -># produced

#7 color(red)(cancel(color(black)("moles CH"_4))) * (color(purple)(2)color(white)(.)"moles H"_2 "O")/(1color(red)(cancel(color(black)("mole CH"_4)))) = "14 moles H"_2"O" -># produced

I recommend checking out this excellent video on how to use mole ratios to solve stoichiometry problems