Question

The number of moles and the mass of oxygen formed by the decomposition of 1.252 g of mercury(II) oxide?

Answer 1

We address the stoichiometric equation....

`HgO(s) + Delta rarr Hg(l) + 1/2O_2(g)uarr`

Explanation:

`HgO(s) + Delta rarr Hg(l) + 1/2O_2(g)uarr`

With respect to `HgO` we decompose a molar quantity of `(1.252*g)/(216.59*g*mol^-1)=5.78xx10^-3*mol`...

And we get HALF this molar quantity of dioxygen gas...

`5.78xx10^-3*molxx32.00*g*mol^-1=0.1850*g`...