(2) 9.06 g(4) 906(3) 0.906 gTwo oxides of a metal contain 72.4% and 70

1.	(2) 9.06 g(4) 906(3) 0.906 gTwo oxides of a metal contain 72.4% and 70% of metal respectively? If formula of 2nd oxide is M203, find thatof the first(2) M,O,(4) MO(1) MOA sample of CaCO, has c
Answer» For Oxygen, we can calculate as w(%) = 100 -70 = 30 But Ar(O) = 16 Based on the law of proportion - [% / Ar], so according to it, = 70/x : 30/16 = 2:3 = 70/x : 1.875 = 2:3, where x = 56 (Fe) So, if w(O) = 100 – 72.4 = 27.6 (%) =72.4/56 : 27.6/16 = 1.2928 : 1.725 = 1 : 1.334 = 3 : 4 The answer should be - M3O4 (Fe3O4) .

(2) 9.06 g(4) 906(3) 0.906 gTwo oxides of a metal contain 72.4% and 70% of metal respectively? If formula of 2nd oxide is M203, find thatof the first(2) M,O,(4) MO(1) MOA sample of CaCO, has c

Answer»

For Oxygen, we can calculate as w(%) = 100 -70 = 30

But Ar(O) = 16

Based on the law of proportion - [% / Ar], so according to it,

= 70/x : 30/16 = 2:3

= 70/x : 1.875 = 2:3, where x = 56 (Fe)

So, if w(O) = 100 – 72.4 = 27.6 (%)

=72.4/56 : 27.6/16 = 1.2928 : 1.725 = 1 : 1.334 = 3 : 4

The answer should be - M3O4 (Fe3O4) .