Question

3.21 gm of a metal was strongly heated in the air which formed 3.64 om of metal
oxide. Calculate the equivalent weight of the metal
8 A metal carbonate was strongly heated in the crucible to a constant weight. The weight​ of metal

Answer 1

3.21g of metal was strongly heated in the air which formed 3.64gm of metal oxide.

2M + (x/2)O2 ⇔M2Ox , here x is valency of metal.

here it is clear that 2 moles of metal form one mole of metal oxide.

so, 1 mole of metal forms 0.5 mole of metal oxides

so, 0.5 × 3.21/M = 3.64/(2M + 16x)

or, 3.21(2M + 16x) = 7.28M

or, 6.42M + 51.36x = 7.28K

or, 7.28M - 6.42M = 51.36x

or, 0.86M = 51.36x

or, M/x = 51.36/0.86 = 59.72 ≈ 60 g

hence equivalent weight of metal is 60.

in second question, data is insufficient. we can't get weight of metal.