The ammonia prepared by treating ammonium sulphate with calcium hydroxide is completely used by NiCl2.6H2O to form a stable coordination compound. Assume that both the reactions are 100% complete. If 1584 g of ammonium sulphate and 952g of NiCl2.6H2O are used in the preparation, the combined weight (in grams) of gypsum and the nickel-ammonia coordination compound thus produced is___. (Atomic weights in g mol–1: H = 1, N = 14, O = 16, S = 32, Cl = 35.5, Ca = 40, Ni = 59) (JEE-ADVANCED 2018 PAPER-2)

According to Stoichiometric equation:

Galena (an ore) is partially oxidized by passing air through it at high temperature. After some time, the passage of air is stopped, but the heating is continued in a closed furnance such that the contents undergo self-reduction. The weight (in kg) of Pb produced per kg of O2 consumed is______ . (Atomic weights in g mol–1 : O = 16, S = 32, Pb = 207). (JEE-Advanced 2018 Paper-2)

We calculate ammount of Pb According to Stoichiometric equation given as

Calculate how much H2SO4 will be obtained from 400 gm of Oleum sample having labelling 104.5%?

SOLUTION: 104.5 % labelled means 100 Oleum sample required 4.5 gm water to completely destroyed free SO3 present in 100 gm sample

100 gm Oleum sample require 4.5 gm water to destroyed all free SO3

Weight of SO₃ in destroyed by 18 gm water is =18/18x80= 80 gm 

Weight of H₂SO₄ = 400-80= 320 gm present in 400 gm Oleum sample

Weight of H₂SO₄ newly formed is =18/18x98= 98 gm

Total Weight H₂SO₄ =320+98=418 gm

A mixture of H2CO3 liquid and CO2gas is labelled as Oleum sample . 50 gm such mixture contains 22% CO2 , find out the % labelling of such mixture.

SOLUTION:

                     Given 22% CO2

 % labeling(X) of CO₂ = 109%

0.5 gm of fume H2SO4 (Oleum ) is diluted with water, this solution is completely neutralised by 26.7 ml of 0.4 N NaOH. Find the percentage free SO3 in sample solution.?

SOLUTION: Given total wt of Oleum sample is 0.5 gm, let x gm SO3 and (0.5-x) H₂SO₄  

                             (E _wt= SO₃=80/2=40 gm and E _wtH₂SO₄=98/2=49 gm)

At equivalent point     

      No of equivalent of SO₃ + No of equivalent H₂SO₄= no of equivalent of NaOH