Magnesium reaction..need some help?

Question

Assume that you start with 4 mole of magnesium metal, determine:

1.The number of grams of MgO(s) that would be produced in the following reaction (enter your answer with 4 significant figures):


Mg(s) + 1/2 O2(g) -----> MgO(s)


cont....


2.The number of grams of Mg3N2(s) that would be produced in the following reaction (enter your answer with 4 significant figures):


Mg(s) + 1/3 N2(g) -----> 1/3 Mg3N2(s)

Answer 1

(I already did the first part in your earlier question)

Part #2

Assuming the same 4 moles of magnesium as before:

The easy way; first determine % weight Mg in Mg3N2

(3 x 34.305)/[(3 x 34.305) +(2 x 14.01)] = 72.92/100.9 x

100% = 72.27% Mg

Now, determine weight of Mg3N2 that can be created from 4 moles of Mg:

4 x 24.305 x (1/0.7227) = 134.5g

You can also see from the equation (if you don't get totally confused by all the subscripts etc) that if we started with 4 moles of Mg we would produce 4/3 mole of Mg3N2 or:

4/3 x [(24.305 x 3) + (14.01 x 2)] = 134.6g