How many grams of water are produced when 2.50 mol oxygen reacts with hydrogen?

Given:

2.5 mol O2

Required:

Grams of water

Solution:

2H2 + O2 -> 2H2O

Moles of H2O = (2.5 mol O2)(2 moles H2O/1 mole O2) = 5 moles H2O

Molar mass of H2O = 18.02 g/mol

Mass of H2O = (5 moles H2O)( 18.02 g/mol)

Mass of H2O =90.1g H2O

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.