Moderators: Chem_Mod, Chem_Admin

Posts: 12
Joined: Fri Sep 29, 2017 7:04 am


Postby venning-1J » Thu Oct 05, 2017 4:59 pm

A reaction vessel contains 5.77 g of white phosphorus and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorous (III) oxide, P406; P4 (s) + 3O2 (g) --> P406(s). IF enough oxygen is present, the oxygen can react further w/ this oxide to produce phosphorus (V0 oxide, P4010; P406 (s) + 2 O2 (g) --> P4010 (s). (a) What is the limiting reactant for the formation of P4010? (b) What mass of P4010 is produced? (c) how many grams of the excess reactant remain in the reaction vessel?

I need help with part b and c.

In part I I found that 0.04657 mol od P4O6 were created and that 0.0406 mol of O2 remain. Since the limiting reactant in P406 (s) + 2 O2 (g) --> P4010 is o2, I figured that at most 0.0406 mol of P4O10 could be created and I multiplied it by its molar mass to find how many grams would remain. I'm still not getting the right answer though.

Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 435 times

Re: M11

Postby Chem_Mod » Thu Oct 05, 2017 6:12 pm

The limiting reactant in the second reaction is indeed O2, but the amount of P4010 you are making is 0.04657 mol divided by 2. To figure that out you have to look at the balanced chemical equation which tells us that P406 : O2 : P4010 are present 1:2:1 ratio! I hope this helps to get the right answer!

Shawn Patel 1I
Posts: 54
Joined: Thu Jul 27, 2017 3:01 am

Re: M11

Postby Shawn Patel 1I » Thu Oct 05, 2017 6:28 pm

You have all your numbers correct, but I think the reason you aren't getting the right answer is because the ratio of the oxygen to the product, phosphorus(V) oxide is 2:1. This means that when calculating for the final product, you have to divide the moles of oxygen remaining by two. This should leave you with the correct amount of product. Then in order to get mass, you multiply by the molar mass.

For part c, you have to find the amount of excess reactant, which can be done by using the 2:1 ratio again. Then you just multiply by the molar mass of the reactant and you should have the amount of excess reactant.

I hope this helps.

Return to “Limiting Reactant Calculations”

Who is online

Users browsing this forum: No registered users and 1 guest