Question

the mass of Cuo obtained by heating of metallic with nitric acid and subsequent ignition was found to be 2.7 gram in another experiment , 1.15 gram of copper oxide on reduction yield 0.92 gram of copper show that data illustrates the law of constant composition​

Answer 1

Answer:

Molecular mass of CaCO3=40+12+(16*3)=100g

So, we don't have to do much in this question!!!

Now, we know ,

CaCO3 (by heating)→CaO + CO2 (↑)

therefore, 1 mole of CaCO3 gives 1 mole of CaO…..

So, 1 mole of CaCO3 = 100g of CaCO3

AND 1 mole of CaO = (40+16)g=56g of CaO

So, 100g of CaCO3 gives 56 g of CaO on heating …..

Thanks for a good question…

Answer 2

Answer:

In the first sample of copper oxide (obtained by the action of nitric acid on Cu), the mass of Cu is 2.16 g and the mass of oxygen is 2.7−2.16=0.54 g.

The ratio of the mass of Cu to the mass of oxygen is

0.54

2.16

=4:1

In the second sample of copper oxide (which reacts with hydrogen) , the mass of Cu is 0.92 g and the mass of oxygen is 1.15−0.92=0.23 g.

The ratio of the mass of Cu to the mass of oxygen is

0.23

0.92

=4:1

Hence, this illustrates the law of definite Proportions.