N2(g) + 3H2-> 2NH3(g) How many grams of ammonia (NH3) can be produced from the reaction of 28.0 g of N2 and 25.0 g of H2?

Question

 Need Chemistry help, I do not understand absolutely nothing!

Answer 1

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.