30 ml of 0.2 N BaCl2 is mixed with 40 ml of 0.3 N Al2(SO4)3 . How many gram of BaSO4 is formed?

Asked by sonali garg | 23rd Oct, 2014, 05:57: PM

Expert Answer:

                                    3 BaCl2   + Al2(SO4)3 →          3BaSO4  + 2AlCl3

mM before reaction        30x0.2         40 x0.3

                                    =6                 =12                     0              0

mM  after reaction           0                  (12-1)                  3               2

 

Since  6mM of BaCl2 reacts with one mole of  Al2(SO4)3

So m M of BaSO4 formed = 3 = wt./ m. wt. x 1000

So Wt. of BaSO4 formed = 3 x 233  = 0.699 g

                                            1000

Answered by Arvind Diwale | 26th Oct, 2014, 02:45: PM

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