Question

Methane CH4 gas and oxygen O2 gas react to form carbon dioxide CO2 gas and water H2O vapor. Suppose you have 11.0 mol of CH4 and 7.0 mol of O2 in a reactor. Suppose as much as possible of the CH4 reacts. How much will be left? Round your answer to the nearest 0.1 mol

Answer 1

CH4 (g)   + 2 O2 (g)    ----------> CO2 (g) +   2 H2O (g)

1                   2                                1                  2         

11.0 mol       7.0 mol      

here O2 is limiting reagent. because moles of O2 is less .

CH4 is excess.

1 mol CH4   ------->   2 mol O2

    ?? mol CH4 ---------> 7 mol O2

moles of CH4 = 7 x 1 / 2 = 3.5 mol

here we have 11 moles of CH4 . so

moles of CH4 left = 11 - 3.5 = 7.5

moles of CH4 left = 7.5 mol