Question

What volume of o2 gas measured at 1atm and 273k will be formed by the action of 100ml of 0.5 n kmno4?

Answer 1

The balanced equation for this reaction is-
        
                                    2KMnO4  ->  K2MnO4 + MnO2 + O2
Molecular weight:        2 * (158)g                                      (16*2)g

1 mole of an ideal gas occupies 22.4l of volume. so here volume of O2 produced is 22.4L.

To solve this we should know,
            
                   1L of 1N KMnO4 contains -> 158/5= 31.6 g of KMnO4
                   1000ml of 1N would have  ->             31.6 g
                   1ml       of 1N would have  ->             31.6g/1000
                   100ml of 1N would have    ->             (31.6g /1000)*100
                                                                            = 3.16g
                    100 ml of 0.5N would have->          = (3.16*0.5) =1.58g

Now, we know that,
                                  (2*158)g of KMnO4 -> 22.4l
                            So,  1.58g     of KMnO4 -> (22.4l/2*158)* 1.58
                                                                       = 0.112l

Ans:- 0.112L 0f O2 is produced.