This is a practiced test question,

Mg3N2 + 6H20 --> 3Mg(OH)2 + 2NH3

How many moles of NH3 are liberated when 2 moles of Mg3N2 is reacted with 9 moles of H20?

Thanks

First, you need to see which reactant is in excess, or which is limiting. Let's see. If you use 2 mol of Mg3N2, you'll need 12 mol of H2O to make all the reactants react. Since you are provided with only 9 mol of H2O, you have Mg3N2 in excess.

Work with the number of moles of water available.

6 moles of water gives 2 moles of NH3
So, how many moles of ammonia would 9 moles of water give off? That's only proportion!

You just need to understand first! :)

Look at it this way. Since you have 2 moles of Mg3N2, write the equation this way:

2Mg3N2 + 12H20 --> 6Mg(OH)2 + 4NH3

Of course, since you only have 9 moles of H2O, you won't have enough to react. Therefore, as the previous answerer stated, Mg3N2 is in excess. So, rewrite it this way:

2*9/12 Mg3N2 + 12 * 9/12 H2O = 6 * 9/12 Mg(OH)2 + 4 * 9/12 NH3

or

1.5 Mg3N2 + 9 H2O = 4.5 Mg(OH)2 + 3 NH3

You'll have 0.5 moles of unreacted Mg3N2 left in the mixture.