Question

HELP NIOS CHEM EXAM IS TOMORROW AND I KNOW NOTHING PLEASE HELP
The reaction 2Al +3MnO Al2O3 + 3Mn proceeds till the limiting reagent is consumed. A mixture of 220g Al and 400g MnO was heated to initiate the reaction.
(a ) Which is the limiting reagent?
(b) Which initial substance is remained in excess and by how much?

Answer 1

For a: First find out the molecular weight of the compounds
Aluminium = 26.982 g/mol
Manganese = 54.938 g/mol
Oxygen = 15.999 g/mol

The reactions work with ratios
2 mole of aluminium reacts with 3 mole manganese (II) oxide to form 1 mole of aluminium oxide and 3 mole manganese
So the ratio is 2:3->1:3
If you know you have 220g Al and 400g MnO
You need to convert those amounts into moles

26.982 gram aluminium is 1 mole
so 220 gram aluminium is 220/26.982 = 8.15 mole Al

1 mole manganese (II) oxide is 70.937 gram (54.938+15.999)
400 gram manganese (II) oxide = 400/70.937 = 5.64 mole MnO

according to the ratio 2 mole Al reacts with 3 mole MnO
You have a total of 8.15 mole Al and 5.64 mole MnO
divide 8.15 by 2 and 5.64 by 3
this will make 4.08 and 1.86

for the amount of aluminium you have the reaction could take place 4 times but with the amount of MnO you have, the reaction can only take place less then 2 times. So the limiting reagent is Manganese (II) oxide.

You now also know that the initial substance aluminium is remained in excess.
To calculate how much Al is left after all the MnO is gone we take a look at the last calculation. Al = 4.08 and MnO=1.86

Now subtract 1.86 from 4.08
You will have 2.22 left. That times 2 because it goes per 2 moles. You'll have 4.44 moles of aluminium left. You multiply that by the molecular weight of aluminum (26.982) and you know you have 119.8 grams aluminum left after the reaction is done.

Hope this helps :)

Answer 2

For the molecular weights you can like on a periodic table like http://www.ptable.com
You will probably get a copy of a ptable on your exam