## Fundamentals Homework Problem M11

Victoria Zheng--2F
Posts: 103
Joined: Fri Aug 09, 2019 12:17 am

### Fundamentals Homework Problem M11

M11: A reaction vessel contains 5.77g of white phosphorus and 5.77g of oxygen. The first reaction to take place is the formation of phosphorus(III) Oxide, P4(s)+3O2(g)-->P4O6(s). If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus(V) Oxide, P4O10: P4O6(s)+2O2(g)--> P4O10(s).
My question is for part (B) What mass of P4O10 is produced? Since oxygen is the limiting reagent, how do I know what amount of oxygen is left from the first equation to react with P4O6?

605379296
Posts: 58
Joined: Sat Aug 17, 2019 12:16 am

### Re: Fundamentals Homework Problem M11

So from part a, we know that there are 4.47 grams of O2. We then take 5.77g of O2- 4.47g O2 to get the excess amount of O2. Then, we get 1.30 grams of O2 and divide that by the molar mass of oxygen; next, you multiply that by the ratio of P4O10 to O2 in the second equation (1 mol P4O10: 2 mol O2) [this step is to first convert the O2 moles into moles of P4O10, so we can later convert the moles into the final molar mass for P4O10]. Lastly, we multiply by the molar mass of P4O10 (283.88 g/mol) to finally get 5.77g of P4O10 produced. In mathematical terms, it's set up like this: (1.30g O2)/(32g/mol O2)*(1 mol P4O10)/(2 mol O2)* (283.88 g/mol P4O10)/ (1 mol P4O10) = 5.77 g P4O10

Victoria Zheng--2F
Posts: 103
Joined: Fri Aug 09, 2019 12:17 am

### Re: Fundamentals Homework Problem M11

Thank you!

MMckinney_4H
Posts: 61
Joined: Sat Sep 28, 2019 12:16 am

### Re: Fundamentals Homework Problem M11

Overall the oxygen is the limiting factor, but for the first reaction, the phosphorous is limiting how much P4O10 is created. With the .04657125 mols of P4, you can make .04657125 mols of P4O6 as the ratio is 1:1, but you only use up .13971375 mols of O2 because the ratio between P4 and O2 in the first reaction is 1:3 (so you multiply the amount in mols of P4 that you have by 3 to get the amount of O2 you need). You then subtract the .13961375 mols of O2 used in the first reaction by the total amount that you had (.18032377) to get the amount still available for the second reaction. I hope this helps :)