Question

Chem 1. Hello! I'm new here and I'd like if someone did this and explained how it was done! Thank you.

What theoretical mass of MgO is produced if 0.150 g Mg3N2 is produced by strongly heating a 2.000 g sample of Mg in air?

Answer 1

Hi!
If we burn Mg in air, we get two possible products: Mg3N2 and MgO, right?
Here are their equations:
3Mg+N2 -> Mg3N2
2Mg+O2 -> 2MgO
We use N2 and O2 instead of N and O because Nitrogen and Oxygen usually exist as N2 and O2 in open air.
So first, we get how many grams of Mg was used up in making 0.150g of Mg3N2:
We see that we need 3 moles of Mg to make 1 mole of Mg3N2.
So first we have to know how many moles of Mg3N2 are there in 0.15g of it and then multiply it to 3 according to the ratio that's provided by the balanced equation.
\[(0.15/100.9494)(3)\]= \[4.458\times10^{-3} moles\]
100.9494g being the molar mass of Mg3N2
Now, we have the appropriate number of moles, we multiply it to the molar mass of Mg to get the grams of Mg used up to make those 0.15g of Mg3N2.
\[(0.00458\times24.305)=0.10834g\]
Now, we subtract this from our initial sample which is 2.0 g of Mg and get 1.89166g of Mg left to react with O2 in the air.
So, like we did before, we get how many moles of Mg is there in 1.89166g of it by dividing it with its molar mass, and once we get that, we multiply it to the molar mass of MgO.
\[(1.89166/24.305) = 0.07783\]
\[(0.07783\times40.3044) = 3.1369g\]
We don't need to multiply it to the number of moles of MgO2 this time because we need 2 moles of Mg to make 2 moles of MgO2. That means, We can make a mole of MgO2 with a mole of Mg. A one to one correspondence.
Therefore, assuming it's a complete reaction, we get 3.1369 g of MgO2 with the situation you presented.