Question M11

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ConnorRazmaDis2I
Posts: 55
Joined: Fri Sep 28, 2018 12:26 am

Question M11

Postby ConnorRazmaDis2I » Sun Oct 07, 2018 8:44 pm

I understand how to do the first reaction P4 + 3O2 = P4O6. I found the limiting reactant to be P4. I was just wondering how I calculate how much oxygen is left for the second reaction

604656370
Posts: 30
Joined: Fri Sep 28, 2018 12:25 am

Re: Question M11

Postby 604656370 » Sun Oct 07, 2018 9:11 pm

The first step woud be to find the number of moles of O2 and P4 in the reaction given that we have 5.77g of each. For example, since the molar mass of O2 is 32g/mol and we have a sample with 5.77g, there will be 5.77g / (32g/mol) = 0.180 mol of O2. Similarly, you can calculate that there are 0.0465 mol of P4. Next, given the balanced equation, we know that 1 mole of P4 reacts with 3 moles of O2 to produce 1 mole of P4O6. Therefore 0.0465 mol of P4 will require 3 X 0.0465 = 0.1395 mol of O2. Since we actually have 0.180 mol of O2, there will be an excess of 0.0405 mol of O2 (0.180 - 0.1395) available for the next reaction. Given this quantity and the moles of P4O6 produced from the first reaction, you can determine the quantity of P4O10 that will be produced. Hopefully this helps!

dgerges 4H
Posts: 65
Joined: Fri Sep 28, 2018 12:24 am

Re: Question M11

Postby dgerges 4H » Wed Oct 10, 2018 8:40 pm

Yes! and in the reaction that creates p4o10 from p4o6 and o2, o2 will be the limiting reactant.


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