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

Apr 13, 2017

Approx. $15 \cdot g$ of magnesium metal.........

#### Explanation:

We need (i) a stoichiometric equation:

$M g \left(s\right) + {O}_{2} \left(g\right) \rightarrow M g O \left(s\right)$,

and (ii), equivalent quantities of the product $M g O$:

$\text{Moles of magnesium oxide} = \frac{25 \cdot g}{40.30 \cdot g \cdot m o {l}^{-} 1} = 0.620 \cdot m o l$.

Now from the stoichiometry of the equation, if there are $0.620 \cdot m o l$ $M g O$, then there MUST have been at least $0.620 \cdot m o l$ $\text{magnesium metal}$, and this quantity represents a MASS of 0.620*molxx24.31*g*mol^-1=??g