Amont of sulphuric acid required to meutralize 10l of ammonia at stp is
Answers
2NH3 + H2SO4 = (NH4)2SO4
Thus 2 moles of ammonia will neutralise 1 mole of pure sulphuric acid.
From the ideal gas equation, PV = nRT, you can deduce that at STP, 1 mole of gas occupies 22.4 litres.
Therefore, 10 litres of NH3 at STP is equal to 10/22.4 moles of NH3. From the above equation the number of moles of sulphuric acid required is one-half of the number of moles of NH3, ie 0.5 x 10/22.4 moles of H2SO4.
The molecular weight expressed in gm is the molarity. Thus the mass of H2S04 is:
0.5 x 10 x 98.08/22.4 g H2SO4 = 21.89 g
In practice conc H2SO4 is at 98% purity so that the mass of H2S04 required will need to be adjusted. To convert to volume assume the SG of the conc acid is 1.84 g/cc from which the volume of the acid is 21.89/1.84 = 11.90 cc. Adjusting for the 98% purity the volume of sulphuric acid will be 11.90 x 100/98 = 12.14 cc.