I we react 52 grams of Magnesium and 52 grams of oxygen, what is made and how much is made?

Question

Use the stoichiometry and the mass to mass relationships to find the answer.
The steps are needed.....

Answer 1

First we need to find the limiting reagent in this reaction.
No of moles of Mg = 52/24 = 2.167
No of moles of O2 = 52/32 = 1.625
Reaction: 2Mg + O2 -> 2MgO
In this case Mg is the limiting reagent (as 2.167mol Mg can only react with a maximum of 1.0835mol of O2 for complete reaction.)
We base the equation on the no of moles of limiting reagent.
So number of moles of MgO formed = 2.167 (by mole ratio of equation)
Mass of MgO (magnesium oxide) formed = 2.167 x (24 + 16)
= 86.68g

Answer 2

Write the reaction:
2Mg +O2 -> 2MgO

Why MgO? Mg is a +2 alkaline earth metals. It forms typically ionic compounds with non-metals. So, with oxygen, which has a nominal -2 position, stoichiometry indicates a 1:1 ratio of Mg to O. The reaction is written as above, since oxygen occurs as the dimer O=O,

Now the mass bit. Which reactant is limiting? For oxygen, we have 52/32 or 1.62 moles of the dimer. For Mg, we have 52/24 or 2.167 g-atoms. The equation indicates that 2 g-atoms of Mg react with 1 g-mole of dimer O2. Mg is limiting, since 2.167 g-atoms require 1.084 g-mole of the dimer and more is available.

So, since the weight of MgO is 40, our 52 g of Mg produces 86.7 g of MgO. The remaining weight is that of the unused O2.