question M.11

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ariaterango_1A
Posts: 57
Joined: Wed Sep 18, 2019 12:15 am

question M.11

Postby ariaterango_1A » Mon Oct 07, 2019 10:44 pm

A reaction vessel contains 5.77g of white phosphorous and 5.77 g of oxygen. The first reaction to take place is the formation of phosphorous (III) oxide, P4O6: P4+ 3O2 -> P4O6. If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorous (v) oxide, P4O10: P4O6 + 2O2 -> P4O10. a) what is the limiting reactant 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 know how to get the limiting reactant of the first reaction I'm just not sure how to use those results to find the limiting reactant of the second equation?

JasonLiu_2J
Posts: 93
Joined: Sat Aug 24, 2019 12:17 am

Re: question M.11

Postby JasonLiu_2J » Mon Oct 07, 2019 10:55 pm

Since the question states that if enough oxygen is present then it will react in the second equation, you would use the remaining (Excess) oxygen you have after the first reaction as the starting amount of oxygen you have for the second reaction. In order to find the amount of oxygen that remains, you would need to determine how much is used up in the first reaction. Since P4 is the limiting reaction in the first equation, all of the phosphorous you had initially is used up. To find the amount of oxygen left, convert the amount of moles of phosphorous used to the moles of oxygen that are used up using the mole ratio of 1:3 phosphorous to oxygen. Then, subtract the moles of oxygen used from the moles of oxygen that you had initially (you would need to convert 5.77 grams of O2 to moles to find how much oxygen you had initially). The resulting value is the initial amount of oxygen you have for the second equation in moles. Now you need the amount of P4O6 (the other reactant in the second equation) that you have initially. Repeat the process that you used to find the amount of oxygen used in the first equation but instead solve for the amount of P4O6 produced based on the limiting reactant. At this point, you would have the amount of P4O6 produced from the first reaction and the amount of O2 remaining after the first equation. These are your initial amounts of the reactants in the second equation. Using these amounts, you can then solve for the limiting reactant in the same way that you found the limiting reactant of the first reaction.
In summary: Determine the amount of O2 remaining and P4O6 produced in the first reaction when the P4 is completely used up (limiting reactant) and use these values as the inital amounts of reactants in the second equation to determine the limiting reaction of the second equation.
Hope this helps!

ariaterango_1A
Posts: 57
Joined: Wed Sep 18, 2019 12:15 am

Re: question M.11

Postby ariaterango_1A » Tue Oct 08, 2019 8:31 am

To find the excess in the first reaction do you compare the molar ratios of the reactants to each other? For instance, .047 moles of P4 * (3 mol of O2/ 1 mol of P4) and .18 moles of O2 * (1mol of P4/ 3 mol of O2) ? From there I found that P4 is limiting and oxygen is in excess so you would subtract .18 mol of O2- .14 mol of P4 to get .04 moles of excess O2? Is this correct for the first part of the equation?

Amy Pham 1D
Posts: 88
Joined: Fri Aug 09, 2019 12:15 am

Re: question M.11

Postby Amy Pham 1D » Tue Oct 08, 2019 4:40 pm

Yes! That is the correct first step to finding the limiting reactant, then you would use the 0.0466 mols P4O6 and compare it with 0.0406 mols O2 you found in the previous step to find which one is the limiting reactant in the second reaction, P4O6 (s) +2 O2 (g) --> P4O10. As it turns out, O2 is the limiting reactant.


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