Question

Are masses of `0.24*g` `Mg` metal and `0.64*g` `O_2` gas stoichiometric with respect to the formation of `MgO`?

Answer 1

`Mg(s) + 1/2O_2(g) rarrMgO(s)`

Explanation:

The stoichiometric reaction shows that `24.3*g` reacts with `16.00*g` of dioxygen gas to give approx. `40*g` of the oxide.....

And for the quoted masses we can interrogate the molar quantities....

`"Moles of magnesium"=(0.24*g)/(24.3*g*mol^-1)=0.01*mol`.

`"Moles of dioxygen"=(0.64*g)/(32.0*g*mol^-1)=0.02*mol`.

And thus a 2:1 molar ratio as required by the stoichiometry.