How many liters of oxygen are needed to exactly react with 25.8 g of methane at STP? #CH_4(g) + 2O_2(g) -> CO_2(g) + 2H_2O(l)#?

1 Answer
May 20, 2018

Answer:

Well at #"STP"# one mole of gas occupies #22.4*L# (this depends on syllabus; luckily, the definition is included as supplementary material on most exam papers....)

Explanation:

And so we gots the combustion equation...

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

And so #"moles of methane"=(25.8*g)/(16.04*g*mol^-1)=1.536*mol#..

And so we require TWICE this molar quantity of dioxygen gas...

And this constitutes a VOLUME of....

#2xx1.536*molxx22.4*L*mol^-1~=70*L#

And why should we study this reaction? Well, it heats our homes, and cooks our dinner, and sometimes drives our cabs. In short it allows us to live beyond 30...