## M11

Hyein Cha 2I
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### M11

A reaction vessel contains 5.77 g of white phosphorous and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorus (III) oxide, P4O6: P4 + 3O2 --> P4O6(s). If enough oxygen is present, the oxygen can react futher with this oxide to produce phosphorus (V) oxide, P4O10: P4O6 + 2O2 --> P4O10(s).

a) What is the limiting reaction for the formation of P4O10?
b)What mass of P4O10 is produced?
c) How many grams of the excess reactant remain in the reaction vessel?

I do have the solutions manuel, yet I still do not understand how to do this question. It sure is a long and tideous question, but can anyone please help me out?

I especially do not understand part A. If someone could please explain part A, then I think I can do part b and c by myself.

Thanks a bunch!
Last edited by Hyein Cha 2I on Thu Oct 05, 2017 5:30 pm, edited 1 time in total.

Kelly Kiremidjian 1C
Posts: 62
Joined: Fri Sep 29, 2017 7:04 am

### Re: M11

I have the same question.Couldn't quite follow the solutions manual

Rachel Wang
Posts: 49
Joined: Fri Sep 29, 2017 7:04 am

### Re: M11

I did it a little differently from the book but got similar answers (around .05 off - not sure if that's a huge problem), so hope this helps c:

You have to find the limiting reactant of P4O6 first.

I converted 5.77g P4 to moles:
The molar mass of P4 is 123.88g
5.77g/123.88g = 0.0466 mol P4.
The ratio of P4 to P4O6 is 1:1 so 0.0466 mol P4O6 would be produced.

I then converted 5.77g O2 to moles:
The molar mass of O2 is 32g.
5.77g/32g = 0.1803 mol O2.
The ratio of O2 to P4O6 is 3:1, so .1803/3 = 0.0601 mol P4O6 would be produced.

0.0466 < 0.0601; P4 is the limiting reactant of P4O6, meaning there's excess O2 and thus it makes sense to carry forward to the second reaction.

To find the limiting reactant of the second equation, you need to find how many moles of O2 are leftover from the first equation.
In the first equation, you used up all 0.0466 mol P4. The ratio of P4 to O2 in the first equation was 1:3, so 0.0466 x 3 = 0.1397mol O2 used.
Subtract the O2 used from the .1803mol you had originally.
0.1803-0.1397=0.0406mol O2 for the second rxn.

Now you have to do the same ratio comparison to see whether O2 or P4O6 limits P4O10.

The ratio of O2 to P4O10 is 2:1, so .0406/2= .0202mol P4O10 would be produced.
The ratio of P4O6 to P4O10 is 1:1, so .0466 mol P4O10 would be produced.

a) O2 is the limiting reactant.

b) Convert .0202mol P4O10 to grams
The molar mass of P4O10 is 283.88g
.0202mol P4O1 x 283.88g = 5.71g P4O10.

c) The excess regent is P4O6 for the second rxn.
You used up all 0.0406 mol O2.
The ratio of O2 to P4O6 is 2:1, so .0406/2 = .0203 P4O6 used.
Subtract .0203 P4O6 from the 0.0466 mol P4O6 you produced from the 1st rxn.
0.0466 - .0203 = .0263 mol x 219.88g (molas mass of P4O6) = 5.78g P4O6 left.

venning-1J
Posts: 12
Joined: Fri Sep 29, 2017 7:04 am

### Re: M11

Thanks!

Hyein Cha 2I
Posts: 103
Joined: Fri Sep 29, 2017 7:05 am
Been upvoted: 1 time

### Re: M11

Rachel Wang wrote:I did it a little differently from the book but got similar answers (around .05 off - not sure if that's a huge problem), so hope this helps c:

You have to find the limiting reactant of P4O6 first.

I converted 5.77g P4 to moles:
The molar mass of P4 is 123.88g
5.77g/123.88g = 0.0466 mol P4.
The ratio of P4 to P4O6 is 1:1 so 0.0466 mol P4O6 would be produced.

I then converted 5.77g O2 to moles:
The molar mass of O2 is 32g.
5.77g/32g = 0.1803 mol O2.
The ratio of O2 to P4O6 is 3:1, so .1803/3 = 0.0601 mol P4O6 would be produced.

0.0466 < 0.0601; P4 is the limiting reactant of P4O6, meaning there's excess O2 and thus it makes sense to carry forward to the second reaction.

To find the limiting reactant of the second equation, you need to find how many moles of O2 are leftover from the first equation.
In the first equation, you used up all 0.0466 mol P4. The ratio of P4 to O2 in the first equation was 1:3, so 0.0466 x 3 = 0.1397mol O2 used.
Subtract the O2 used from the .1803mol you had originally.
0.1803-0.1397=0.0406mol O2 for the second rxn.

Now you have to do the same ratio comparison to see whether O2 or P4O6 limits P4O10.

The ratio of O2 to P4O10 is 2:1, so .0406/2= .0202mol P4O10 would be produced.
The ratio of P4O6 to P4O10 is 1:1, so .0466 mol P4O10 would be produced.

a) O2 is the limiting reactant.

b) Convert .0202mol P4O10 to grams
The molar mass of P4O10 is 283.88g
.0202mol P4O1 x 283.88g = 5.71g P4O10.

c) The excess regent is P4O6 for the second rxn.
You used up all 0.0406 mol O2.
The ratio of O2 to P4O6 is 2:1, so .0406/2 = .0203 P4O6 used.
Subtract .0203 P4O6 from the 0.0466 mol P4O6 you produced from the 1st rxn.
0.0466 - .0203 = .0263 mol x 219.88g (molas mass of P4O6) = 5.78g P4O6 left.

OMG BLESS YOU!! THIS MAKES SOOO MUCH MORE SENSE TO ME. The book solutions were not rlly making sense to me haha.

Thank you soooo much!!

Sabrina Dunbar 1I
Posts: 57
Joined: Fri Sep 29, 2017 7:07 am
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### Re: M11

I didn't use math to get O2 as the excess reactant, I just reasoned it out because the problem saw that if there is enough oxygen, the process can be further carried out in a secondary reaction, so thank you very much for elaborating on that! I would have been so lost on how to solve for it specifically