How do we work out the limiting reagent in combustion reactions?

1 Answer
Jul 6, 2017

Answer:

You will have to supply a few details..........

Explanation:

Let's take the combustion of methane........which reaction cooked my dinner tonight.

#CH_4(g) + 2O_2(g) rarr CO_2(g) + 2H_2O(g)#

And suppose I used a mass of #10*g# of methane...........Now clearly methane is the LIMITING REAGENT, and the dioxygen was in excess. Do you agree?

#"Moles of methane"=(10*g)/(16.01*g*mol^-1)=0.625*mol#

And we require #0.625*molxx2=1.250*mol# #"dioxygen gas"#, which represents a mass of how much?