Question

Hey again Matt! I have another question if you can help me!

How many grams of magnesium oxide can be produced when 97.2 g Mg react with 88.5 g O2?

The unbalanced equation is: Mg + O2 → MgO

I already balanced the equation to 2Mg + O2 → 2MgO

Answer 1

Hey,
This is an easy one once you get a grip on how it goes.

First obviously balance the equation which you already did!

Then you would take the 97.2 g of Mg and divide it by 24.3g/mol which is the molar mass of Mg!

Answer 2

Ok So I got that and then if we look at the balanced ratio come out that 2 mol Mg = 2 mol MgO which would also mean that 4 mol Mg= 4 mol Mgo?!

Answer 3

Exactly! You got this! Then you would take the 88.5 g of O2 and divide it by its molar mass which is 32 g/mol.

Answer 4

That equals 2.766 mol O2 right?

Answer 5

Perfect Yes! Then do the ratio again comparing moles of MgO to that of O2. 4:2 = 2:1! So 1 mol O2 = 2 mol MgO so 2.766 mol O2 = ??

Answer 6

5.532 mol??

Answer 7

Exactly!

Answer 8

So that means that Mg is the limiting so 4 mol MgO are formed?

Answer 9

Yep and finally you have to do a little multiplication with (4 mol MgO)* (40.3g/mol MgO) which equals??

Answer 10

161.2 g MgO?

Answer 11

Yay!! You figured it out. See not so hard!!

Answer 12

Thanks again Matt I understand it so much better now ☺ So if my final answer looks like this I should be ok right?

2Mg + O2 --> 2MgO
97.2g Mg / (24.3g/mol Mg) =4mol Mg
2mol Mg-> 2mol MgO so 4mol Mg = 4mol MgO
88.5g O2 / (32g/mol O2) = 2.766mol O2
1mol O2 -> 2mol MgO so 2.766mol O2 = 5.532mol MgO
Mg is limiting so 4mol MgO are formed
4mol MgO * (40.3g/mol MgO) = 161.2 g MgO

Answer 13

Absolutely! It's all the steps!