Question #6811e

1 Answer
Apr 5, 2017

Answer:

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#