Aneesh C. answered • 01/16/21
Credentialed Math Teacher and Academic Math Tutor
Hi Gena!
I'm assuming that by MG you mean Magnesium, which is abbreviated "Mg" in the periodic table. If that is the case, then here's how I would approach this problem:
1) First, we need to find the limiting reactant and use that particular reactant's mass to figure out how much product is produced.
The equation for this problem is as follows: 2 Mg + O2 --> 2 MgO
In order to find the limiting reactant, I will figure out how much O2 is needed to fully react with the 17.4 g Mg (you can also figure out how much Mg is needed to fully react with the 450 g O2, but I choose to do the former).
17.4 g Mg x (1 mol Mg / 24.3 g Mg) x (1 mol O2 / 2 mol Mg) x (32.0 g O2 / 1 mol O2)
= 11.46 g O2 needed to fully react with all of the Mg
We actually have 450 g O2, which is clearly more than the 11.46 g O2 needed, so Mg is the limiting reactant
2) Knowing that Mg is the limiting reactant now, we will use its quantity to calculate the amount of MgO produced:
17.4 g Mg x (1 mol Mg / 24.3 g Mg) x (2 mol MgO / 2 mol Mg) x (40.3 g MgO / 1 mol MgO)
= 28.86 g MgO produced
