# Question 6811e

Apr 5, 2017

We have the stoichiometric equation. We would get approx. $4 \cdot g$ of oxide.........

#### Explanation:

$M g \left(s\right) + \frac{1}{2} {O}_{2} \rightarrow M g O$

(I tend to use non-integral coefficents when I deal with bimolecular reagents. Why? Because it makes the arithmetic easier. Of course I do not propose that you have half a molecule of oxygen, but you certainly can have $8 \cdot g$ of the stuff.)

$\text{Moles of metal}$ $=$ $\frac{2.51 \cdot g}{24.3 \cdot g \cdot m o {l}^{-} 1} = 0.103 \cdot m o l$

Given the equation, AT MOST I can form $0.103 \cdot m o l$ of magnesium oxide. Does you agree?

And thus, max. mass of oxide is equivalent to...........

0.103*molxx40.3*g*mol^-1=??*g#