## Fundamentals Problem M.11

Kate_Santoso_4F
Posts: 72
Joined: Fri Sep 28, 2018 12:29 am

### Fundamentals Problem M.11

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, P4O6: 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). (c) How may grams of excess reactant remain in the reaction vessel?
Could someone please explain how to solve part c? I found O2 to be the limiting reactant for the formation of P4O10. Where should I go from there?

Seohyun Park 1L
Posts: 73
Joined: Fri Sep 28, 2018 12:29 am

### Re: Fundamentals Problem M.11

You basically need to find the amount of P4O6 produced from the first reaction. Then, you would find out how much P4O6 is consumed with the O2 in the second reaction. Subtract the P4O6 consumed from the P4O6 used in the first reaction, and you would have the mass of the excess reactant.

Hai-Lin Yeh 1J
Posts: 89
Joined: Fri Sep 28, 2018 12:16 am
Been upvoted: 1 time

### Re: Fundamentals Problem M.11

So, you know that in the second reaction, O2 is the limiting reactant. That means in the second reaction, the excess reactant (the one that there is too much of left) is going to P4O6 and they are asking us how much of that will be left. Remember that in reaction one (the first reaction), the product (P4O6) is going to be used in the second reaction IF there is enough oxygen.

Therefore, since O2 is the limiting reactant, we have to calculate how much P4O6 is actually going to be used:
1.3 g of O2 or 0.0406 mol O2 will be left from reaction 1.

Use mole ratios to determine amount of P4O6 needed:
0.0406 mol O2 ( 1 mol P4O6 / 2 mol O2) = 0.0203 mol P4O6

Convert to grams:
0.0203 mol P4O6 ( 219.89 g / 1 mol P4O6) = 4.47 g
Thus, we know that in the second reaction, 4.47 g of P4O6 will be used up. But, how much P4O6 is produced in the first reaction?

Calculate based off limiting reactant in reaction 1:
0.0466 mol P4 ( 1 mol P4O6 / 1 mol P4) = 0.0466 mol P4O6 ( 219.89 g / 1 mol P4O6) = 10.2 g P4O6

Now, we know that 10.2 g P4O6 is produced. But we only use 4.46 g of it in the second reaction, so subtract to find out how much excess is left:
10.2 - 4.47 = 5.7 g P4O6

Hope this helps!