Question

How much `"copper(II) oxide"` will be obtained from a `12.7*g` mass of copper metal?

Answer 1

Approx....`16*g` of oxide......

Explanation:

We need (i) a stoichiometric equation....

`Cu(s) + 1/2O_2(g) rarr CuO(s)`

And (ii) we need equivalent quantities of metal....

`"Moles of copper metal"=(12.7*g)/(63.55*g*mol^-1)=0.200*mol`

And so upon oxidation, I get `0.200*mol` with respect to `"copper(II) oxide"`, a mass of....

`0.200*molxx79.54*g*mol^-1=??*g`