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`?

Explanation

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`.