## M.11

Sunjum Singh 1I
Posts: 30
Joined: Fri Apr 06, 2018 11:05 am

### M.11

What are the steps in solving for part b and c of number M.11?
Attachments ErinKim1I
Posts: 31
Joined: Fri Apr 06, 2018 11:03 am

### Re: M.11

(b) take the number of moles of oxygen gas you initially have and subtract it by the moles of O2 used in the first equation to produce P4O6.
you should have about .0405 mol O2 for the second equation.
Use that number to calculate how many grams of P4O10 you would be able to produce in the second equation, as O2 is the limiting reactant for the second reaction.

(c) take the number of moles O2 you have at the beginning of the second reaction and use that to find how many moles of P4O6 you would use up in the second reaction. That's the number of mols P4O6 used. Then, take the number of moles P4O6 you have at the beginning of the second reaction and subtract that by the moles P4O6 used. What is remaining is what is left in the vessel.

Sarah Brecher 1I
Posts: 31
Joined: Fri Apr 06, 2018 11:02 am

### Re: M.11

How do you know that O2 is the limiting reactant for part a? After solving, I know that P4 is the limiting reactant for the first equation so do you use the leftover amount of O2 to find the limiting reactant for the second equation? Or do you use the 5.77 g O2 from the beginning to find the limiting reactant?

Chem_Mod
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### Re: M.11

You have to find out how much of the oxygen would be left over (if any is left over) for the next reaction. You can only tell this by explicitly doing the calculations.

Ashley Kim
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### Re: M.11

I thought of a comprehensive solution!
First, find the moles of P4O6 formed for each reactant. The substance that makes the lowest amount is the LM. With that value, you now know how many moles of P4O6 you have for the next reaction.

You still need to figure out how much O2 is left for the next reaction, so you find the moles of O2 used in the first reaction by doing a mole ratio with the moles of product. You then use the original value of 5.77 g/mol to find how many moles of O2 you originally had.

You subtract these two to find the amount of O2 left for the second reaction, and then you use the value of P4O6 formed. Then it's another LM calculation to calculate the moles of product formed from both reactants. You eventually find that O2 is the LM for this reaction, as expected.

Does this make it easier to understand?

105085381
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Joined: Fri Sep 28, 2018 12:15 am

### Re: M.11

Is O2 or P4 the limiting reactant for part a? Can someone please explain which and why. Thanks so much!

ryanhon2H
Posts: 60
Joined: Fri Sep 28, 2018 12:28 am

### Re: M.11

O2 is the limiting reactant for part a.

You first have to figure out the limiting reactant for the first reaction using mole ratios, which turns out to be P4. This means that all of the P4 is used in the first reaction, but only some of the O2 is used.

Since you now know that P4 is the limiting reactant for the first reaction, you can use the moles of P4 to calculate how many moles of the product P4O6 is produced and how much O2 is used up in the first reaction. The moles of P4O6 then becomes a reactant for the second reaction.

Then, subtract the moles of O2 used up in the first reaction from the initial moles of O2, which was calculated using the original value of 5.77g. The result is how many moles of O2 are unused in the first reaction, and is available to be used in the second reaction to produce P4O10.

Use mole ratios to compare the moles of O2 remaining and P4O6 to determine the limiting reactant for the second equation, which turns out to be O2.

Sean Reyes 1J
Posts: 67
Joined: Fri Sep 28, 2018 12:24 am

### Re: M.11

The limiting reactant overall for BOTH reactions together is O2.

Using only the first chemical reaction may make it seem as though the P4 is the limiting reactant, but when looking at both reactions in the formation of P4O10, O2 is found to be the limiting reactant.

Given that both white phosphorus (P4) and oxygen gas (O2) are found in equal masses, you must use their molar masses to find out how many moles of each substance there are. Using the limiting reactant FOR THIS PARTICULAR REACTION, you can then find out how many moles of P4O6 there are and how many moles of O2 are in excess.
Using this excess of O2 for the following reaction reveals why O2 is the limiting reactant in the formation of P4O10; there are less moles of O2 than there are moles of P4O6.

Sean Reyes 1J
Posts: 67
Joined: Fri Sep 28, 2018 12:24 am

### Re: M.11

I just realized I didn't even answer the question my bad hhhhhh.

But using the moles found from part a of O2 and of P4O6, you would convert both to moles of P4O10 using stoichiometry. The lower amount is the maximum amount that you would theoretically be able to create in a lab experiment. Then, using that amount of moles of P4O10, you would find the molar mass of P4O10 and multiply that by the moles to find the grams of P4O10 that you could possibly create.

For part c, you would convert the amount of moles of O2 in the second reaction into moles of P4O6. This amount represents the amount of moles that will react with the O2. Finding the difference between this amount and the original amount at the beginning of the reaction will yield the excess moles of P4O6, which can be converted into grams using molar mass.

Okay NOW I think I'm done. Sorry for not even reading the question ahhhhhh!!!