Limiting Reactant in Multi-Step Reaction

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Tanvi Mamtora 1I
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Limiting Reactant in Multi-Step Reaction

Postby Tanvi Mamtora 1I » Thu Oct 16, 2014 4:14 pm

*This question is based off of the book question M.7 in the Review Section*

In this question it is asked to find a limiting reactant from a series of 2 equations. How do we determine which equation to focus on and how do you convert from one unit of the equation to another on part a? Then on part b what amount would we use to determine the amount of product used in the reaction? Would we simply use the masses of the Limiting Reactant originally given to us or would we have to use some other quantity to convert through the equations? I looked in the solution manual and the mass 1.30 grams was used instead of the original mass of the limiting reactant given, which caused me some confusion.

Thank you in advance for your help! :)

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Re: Limiting Reactant in Multi-Step Reaction

Postby Chem_Mod » Fri Oct 17, 2014 10:05 pm

a) Easiest thing you can do is to add both equations together:
(I)P4 + 3 O2 => P4O6
(II)P4O6 + 2 O2 => P4O10

P4 + 3 O2 + P4O6 + 2 O2 => P4O6 + P4O10
Here, I'm putting everything into one big equation by putting everything that's on the left side of I,II on the left side and the same for the right.

Cancelling and combining like terms has:
P4 + 5 O2 => P4O10

Then you would proceed to determine the Limiting Reagent by comparing:
moles of P4 vs moles of O2 times 5(stoichiometric coefficient)

b)For this question, you solve for the products in sequence, just as how it is presented. The first step uses all of the phosphorous while only using a bit of the oxygen, creating 10.24 g of P4O6.

In the next reaction, the amount you have for P4O6 and O2 must be determined from the previous reaction. In this case, the previous reaction yielded 10.24 g of P4O6 while consuming 4.47 g of O2 leaving you with 5.77-4.47 = 1.30g of O2.

So now the 2nd reaction has, on the left hand side, 10.24 g of P4O6 and 1.30g of O2. Ignore the rest and do the problem as if you just had only these numbers.
P4O6 + 2 O2 => P4O10
10.24g 1.30g

In this reaction, the limiting reagent is O2 and you would approach this like any other problem.
1. Balance Equation
2. Convert to moles
3. Determine ratios using stoichiometric coefficients
4. Determine Limiting Reagent
5. Solve

c)This part is just clean up. Knowing that the 2nd reaction did not use all of the P4O6, you must determine how much is left. Everything from the 1st reaction is transferred over to the 2nd so no need to consider that part. Here you initially have 10.24 g of P4O6 and ended up using 4.47 g, so 10.24-4.47 = 5.7 g. The 4.47 g is calculated by determining how much P4O6 you used. Here, there is a 1:2 ratio between P4O6 and O2. So determine the number of moles of O2 you used knowing:
P4O6 + 2 O2 => P4O10
10.24g 1.30g

nO2 = 1.30 g /(mmass of O2)
from 1:2 ratio, we have: nO2 = 2 x nP4O6 - solve for P4O6
massP4O6 = P4O6 x mmass of P4O6

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