Question

In an experiment 1.288 gram of copper oxide was obtained from 1.03 gram of copper. in other experiment 3.672 gram of copper oxide give on reaction 2.938 gram copper. verify the law of constant proportion.​

Answer 1

First experiment

Copper oxide = 1.288 g

Copper left = 1.03 g

Oxygen present = 1.288 - 1.03 = 0.258 g

 

Percentage of oxygen in CuO = (0.258 × 100) / 1.288 = 19.98 ~ 20 %

 

Second experiment

Copper oxide = 3.672 g

Copper left = 2.938 g

Oxygen present = 3.672 - 2.938 = 0.734 g

 

Percentage of oxygen in CuO = (0.734 × 100) / 3.672 = 20.03 ~ 20 %

So, it is verified that these figures verify the law of constant proportion.

Best wishes!

Answer 2

Explanation:

Cu+O

2

→2CuO

1.03g 0.258g 1.288g

Therefore, since Mass of reactant=Mass of product

It is in accordance with law of conservation of Mass.

So,

Ratio of Oxygen

Ratio of Copper

=

0.258

1.03

=

1

3.99

=4:1