1.179 gram of copper on heating in air give 1.476 gram of black oxide of copper. 1.375 of black copper oxide on reduction with hydrogen gave 1.098 gram of copper show that these results illustrate the law of constant proportion?
Answers
Explanation:
In the first experiment:-
Weight of copper oxide, (CuO)=1.288g
Weight of copper =1.03g
Weight of oxygen = Wt. of copper oxide − Wt. of copper =1.288−1.03=0.258g
∴
Wt. of oxygen
Wt. of copper
=
0.258
1.03
≈
1
4
In the second expriment:-
Weight of copper oxide (CuO)=3.672g
Weight of copper =2.938g
Weight of oxygen = Wt. of copper oxide − Wt. of copper =3.672−2.938=0.734g
∴
Wt. of oxygen
Wt. of copper
=
0.734
2.938
≈
1
4
The proportion of elements of copper and oxygen in the reactions is same.
Hence law of constant proportion is proved.
Answer:
2.42 gms of copper gives 3.025 gms of black oxide(CuO)
=> percentage of copper in cuo = 2.42/3.025 x 100 = 80%
=> percentage of oxygen in cuo = 100 - 80 = 20 %
Ratio of copper : oxygen = 80 : 20 = 4:1
Similarly ,
6.49 gms of black oxide gave 5.192 gm of copper
=> Percentage of copper in cuo = 5.192/6.49 x 100 = 80 %
percentage of oxygen in cuo = 100 - 80 = 20 %
Ratio of copper : oxygen = 80 : 20 = 4:1
Hence in both the reactions the ratio of copper : oxygen is same
Hence these figures are in accordance with law of constant proportion