Dry hydrogen was passed over 1.58 g of red hot copper oxide till all of it completely reduced to 1.26 g of copper (Cu). If in this process 0.36 g of H_2 O is formed, what will be the equivalent weight of Cu and O?

Dry hydrogen was passed over 1.58 g of red hot copper oxide till all o

1.	Dry hydrogen was passed over 1.58 g of red hot copper oxide till all of it completely reduced to 1.26 g of copper (Cu). If in this process 0.36 g of H_2 O is formed, what will be the equivalent weight of Cu and O?
Answer» Solution :Copper OXIDE `+H_2 to Cu + H_2 O` `therefore` equivalent of copper oxide `=` equivalent of Cu `=` equivalent of `H_2 O`. `therefore ("weight of copper oxide")/("eq. weight of copper oxide")=("weight of Cu")/("eq. wt. of Cu")=("weight of" H_2 O)/("eq. wt. of" H_2 O)` Suppose eq. wt. of `Cu and O` are `X and y` RESPECTIVELY and SINCE, eq. wt. of H is 1, we have, equivalent weight of copper oxide `=x+y` equivalent weight of copper `=x` equivalent weight of `H_2 O = 1+y ( E_(H_2 O) = E_(H) + E_(O) )` `therefore (1.58)/( x+y) = (1.26)/( x)= (0.36)/( 1+y)` `therefore x=31.5` and `y=8`.