M.11 hw prob

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Doris Cho 1D
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Joined: Thu Jul 11, 2019 12:16 am

M.11 hw prob

Postby Doris Cho 1D » Mon Oct 07, 2019 11:23 pm

thought I knew how to start this problem, looked at the solutions manual and was very wrong. How do you know to start looking for the mass of oxygen from the first reaction if it's asking for the limiting reactant of the second formation (P4O10)? Are you even supposed to do anything with the P4O10?
why is O2 the limiting reactant?

805394719
Posts: 104
Joined: Wed Sep 11, 2019 12:16 am

Re: M.11 hw prob

Postby 805394719 » Tue Oct 08, 2019 12:55 am

So, the question states that there are two reactions occurring. The first reaction occurs between P4 and O2 to yield P4O6. If enough oxygen is present, the oxygen can react further with this oxide to produce P4O10. First, you need to determine the limiting reactant for the first equation. It states that 5.77g of phosphorus and 5.77g of oxygen were reacted. To begin with, we need to calculate the amount of phosphorus and oxygen in moles and determine which one will be in excess. There are 0.046 moles of phosphorus, and 0.18 moles of oxygen.

The equation is this:
P4 + 3O2 --> P4O6

Looking at the stoichiometric coefficients, we determine that the phosphorus will be completely used and therefore will be the limiting reactant. 0.15 moles of oxygen will be spent, and only 0.3 moles will remain at the end of this reaction.

The second equation is this:
P4O6 + 2O2 --> P4O10

There will be 0.046 moles of P4O6 found in the first equation since the coefficients of P4 and P4O6 are the same.
When 0.046 moles of P4O6 reacts with 0.3 moles of O2, the limiting reactant will be P4O6, and there will be 0.2 moles of oxygen left at the end of the reaction. This excess oxygen will be converted into grams, which makes 3.2g.


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