Question

How much magnesium metal is needed to prepare a `25*g` mass of magnesium oxide?

Answer 1

Approx. `15*g` of magnesium metal.........

Explanation:

We need (i) a stoichiometric equation:

`Mg(s) + O_2(g) rarr MgO(s)`,

and (ii), equivalent quantities of the product `MgO`:

`"Moles of magnesium oxide"=(25*g)/(40.30*g*mol^-1)=0.620*mol`.

Now from the stoichiometry of the equation, if there are `0.620*mol` `MgO`, then there MUST have been at least `0.620*mol` `"magnesium metal"`, and this quantity represents a MASS of `0.620*molxx24.31*g*mol^-1=??g`