Problem M11 part a

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Isabelle Bautista 3H
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Joined: Fri Sep 29, 2017 7:06 am

Problem M11 part a

Postby Isabelle Bautista 3H » Wed Oct 04, 2017 3:47 pm

Here is the question from the textbook (Fundamentals Section M)-
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 phosphorus (lll) oxide, P4O6: P4+ 3O2--> P4O6. If enough oxygen is present, the oxygen can react further with this oxide to produce phosphorus (V) oxide, P4O10: P4O6 + 2O2 --> P4O10.

(a) What is the limiting reactant for the formation of P4O10?

Can anyone explain why, in order to find the limiting reactant of the second equation, the grams of P4 is used, rather than the grams of P4O6 (which would have been created by the first reaction)? I attempted to solve this by determining the LR of the first equation, then using that to determine the amount of P4O6 that would be produced, then using that value to determine the LR of the second equation. Is this because the amount of P4O6 is a theoretical amount, therefore can't be used in actual calculation, so we must use the given amount of P4 even though P4 isn't an explicit reactant in the second equation?

Caroline C 1G
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Joined: Fri Sep 29, 2017 7:05 am

Re: Problem M11 part a

Postby Caroline C 1G » Wed Oct 04, 2017 4:13 pm

By using P4, you eventually discover that oxygen is the limiting reactant of the second equation. This means that oxygen controls the amount of P4O10 produced, not P4O6.

Sarah_Stay_1D
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Re: Problem M11 part a

Postby Sarah_Stay_1D » Thu Oct 05, 2017 2:51 pm

I had trouble with this problem too. First have to use the grams of P4 (which you will find to be the limiting reactant of the first equation), to find the grams of P4O6 (the product of the first reaction). You use the grams of P4O6, and the grams of excess O2 (from the first reaction) to find the limiting reactant in the second reaction.

Eryn Wilkinson 3H
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Joined: Fri Sep 29, 2017 7:07 am

Re: Problem M11 part a

Postby Eryn Wilkinson 3H » Thu Oct 05, 2017 3:24 pm

Is there a way to combine the two reactions and solve from there just using molar ratios? I attempted that way but got it wrong so just wondering if anyone accomplished it?

Isita Tripathi 2E
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Re: Problem M11 part a

Postby Isita Tripathi 2E » Thu Oct 05, 2017 4:17 pm

I think the problem with combining both equations is that you end up deciding between P4 or O2 as the limiting reactants for the production of P4O10, when really you should be deciding between P4O6 and O2. Also, you assume that all of the given amount of oxygen is used in the production of P4O10, so when you do your calculation to find the amount of product, you are using a much larger number (5.77 g instead of 1.30 g), leading to much more product.
My TA said you can use the combined reaction to verify that oxygen is in fact the limiting reactant in the formation of P4O10, but you have to go through both reactions individually to find the correct amount of product.

Alyssa Parry Disc 1H
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Joined: Sat Jul 22, 2017 3:01 am

Re: Problem M11 part a

Postby Alyssa Parry Disc 1H » Thu Oct 05, 2017 7:37 pm

The way I solved this problem was finding the limiting reactant in the first reaction (P4) and then using the leftover O2 and the amount of P406 as the amounts given for the second reactant and then I was able to find the limiting reactant for the formation of P4O10. I did this because the question states that if there is enough oxygen present, the oxygen can react further with P4O6 to produce P4O10.


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