Gr. 11 chemistry finding amount of moles?

Question

i dont get this question please help me :

calculate the amount of product formed (in moles) for each reaction

a) 4P + 5 O2 -> 2 P2O5

4 P= 14.5 mol

5 O2= 18.0 mol

b) 2Al + 3 C2 -> 2 AlCl3

2 Al= 3.6 mol

3 C2= 5.3 mol

could u explain the steps to do this problem thank you

Answer 1

The numbers in front of the chemical elements function as an equivalent of the reaction.

a>____________4P_+_5O2_---> 2P2O5

____before---->_14.5__18.0__________

____reaction->_14.4__18.0_____7.2___ ---> (4 : 5 : 2)

____after------>_0.1____0______7.2___

product: P = 0.1 moles, P2O5 = 7.2 moles

b>____________2Al_+_3C2_---> 2AlCl3

____before---->__3.6___5.3__________

____reaction->__3.53__5.3_____3.53__ ---> (2 : 3 : 2)

____after------>_0.067____0_____3.53__

product: Al = 0.067 moles, Al2Cl3 = 3.53 moles

Sorry for bad english..

Answer 2

It is a little confusing how you have written the moles above. I am going to assume that you have 14.5 moles in total of P and 18.0 moles in total of O2. If I am wrong you can still apply the method.

a) The equation tells you that 4 moles of P react with 5 moles of O2 to give you 2 moles of P2O5

The first thing to do is work out which is the limiting reagent:

The limiting reagent is the reagent that will be completely used up if the reaction goes to completion.

You have 14.5 moles of P.

The amount of O2 you need to react with this amount of P is

5/4 x 14.5 moles (because 4P needs 5O2 to react)

= 18.125 mol

You have only been given 18.0 mol of O2, this means that you do not have enough O2 to react with all the P present. Therefore the O2 is the limiting reagent and the P is said to be in excess

Now, since O2 is the limiting reagent, the maximum product that can form is from reaction of 18.0 mol of O2

5 moles of O2 react to give 2 moles of P2O5

Therefore 1 mole of O2 will gice 2/5 moles of P2O5

So moles of P2O5 produced = 2/5 x 18.0

= 7.2 moles of P2O5

b)

determine limiting reagent:

2 moles of Al are required to react with 3 moles of Cl2

Therefore 3.6 moles of Al will need

3/2 x 3.6 moles of Cl2

= 5.4 moles of Cl2 required to react with 3.6 moles of Al

You only have 5.3 moles of Cl2, therefore not enough Cl2 to react with all the Al, so Cl2 is limiting.

Max product will be from reaction of all Cl2 (because it is the limiting reagent)

3 moles of Cl2 produce 2 moles of AlCl3

Therefore 5.3 mole of Cl2 will give:

2/3 x 5.3 moles of AlCl3

= 3.53 moles of AlCl3