Question

Question #6811e

Answer 1

We have the stoichiometric equation. We would get approx. `4*g` of oxide.........

Explanation:

`Mg(s) + 1/2O_2 rarr MgO`

(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*g` of the stuff.)

`"Moles of metal"` `=` `(2.51*g)/(24.3*g*mol^-1)=0.103*mol`

Given the equation, AT MOST I can form `0.103*mol` of magnesium oxide. Does you agree?

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

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