79. Methane, the chief component of natural gas, burns in oxygen according to the equation
CH,(g) + 2 0,(g) » CO,(g) + 2 H2O(g)
If 7.0 mol of CH, react with an unlimited amount of oxygen, what amount (in moles) of
H2O(g) will be produced?

14 mol

Explanation:

                                CH4(g) + 2O2(g) » CO2(g) + 2H2O(g)

from reaction          1 mol                                                2 mol

given                       7 mol                                                x mol

x = (7*2)/1 = 14 mol

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.