Ammonium chloride, ammonium sulfate and ammonium nitrate are used in fertilisers. Ammonia reacts with nitric acid to make ammonium nitrate.
NH3 + HNO3 → NH4NO3
Calculate the mass of ammonia required to make 5.0 g of ammonium nitrate.
relative formula mass of NH3 = 17
relative formula mass of NH4NO3 = 80

1.0625 g

Explanation:

Number of moles in 5g of NH4NO3 = 5.0g/80g/mol = 0.0625 moles

Again the balanced reaction equation is;

NH3 + HNO3 → NH4NO3

So,

1 mole of NH3 yields 1 mole of NH4NO3

x moles of NH3 yields 0.0625 moles of NH4NO3

x = 0.0625 moles of NH3

Mass of NH3 = number of moles * molar mass

Mass of NH3 = 0.0625 moles * 17g/mol

Mass of NH3 = 1.0625 g

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.