Question

How can we calculate volume of NH3 produced at S.T.P when 200 g NH4Cl is heated with 200g of Ca(OH)2 according to following relation
2NH4Cl + Ca(OH)2 = CaCl2 + 2NH3 + 2H2O​

Answer 1

Explanation:

2NH₄Cl + Ca(OH)₂  →  2NH₃ + CaCl₂ + 2H₂O

Now 2 moles of ammonia is produced from 2 moles of ammonium chloride and 1 mole of calcium hydroxide.

Moles of ammonium chloride of = 100 / 53.5 = 1.86

Moles of calcium hydroxide = 100/ 74 =  = 1.35

We need to find limiting reagent.

1 mole of cal. hydr. reacts with 2 moles of ammo. chl.

So 1.35 moles of cal.hyd. requires 1.35 x 2 = 2.7 moles of ammo.chl.

which is not accessible.

Hence Calcium hydroxide is not limiting reagent.

2 moles of ammonium chloride reacts with 1 mole of calcium hydroxide

So 1.86 of amm.chl. requires 1.86/2 = 0.93 moles of cal.hydr.

, which is available.

So ammonium chloride  is limiting reagent.

Unreacted moles of calcium hydroxide = 1.35 - 0.93 = 0.42 moles

Now, 2 moles of ammonia are formed from 2 moles of ammonium chloride.

Therefore 1.86 moles would be formed from 1.86 moles of ammonium chloride .

Weight of ammonia produced = 1.86 x 17 = 31.62 gm

1 mole of ammonia measures 22.4 L at STP

1.86 mole would measure 1.86 x 22.4 = 41.664 L at STP.