[Mr-E]

Desperately need to understand this asap

4.40g of P4O6 and 3.00g of I2 are mixed and allowed to react according to the equation:

5P4O6 + 8I2 ---> 4P2I4 + 3P4O10

Questions
Q: which reactant is in excess and by what mass?

A: P4O6 by 2.78g

I get to the point of knowing the the moles of each but then have no idea about the ratio.

P4O6 moles = .0201
8I2 moles  =  .0118

Someone said to take .0201 divide by 5 then times by 8 = .0321
then .0118 divide by 8 then times by 5 = .0073

0.02                  0.032
0.007                0.0118
excess = 0.007
limiting = 0.0118

I found this extremely confusing. Could someone explain this or help me out with your knowledge. Would be appreciated. Thanks

[AWK]

P4O6 moles = .0201
8I2 moles  =  .0118

change moles into masses and find the reagent in excess (compare to masses of 5P₄O₆ and 8I₂.

[Nisarg Shah]

I'll go along with your question -

4.40g of P₄O₆ and 3.00g of I₂ are mixed and allowed to react according to the equation:

5P₄O₆ + 8I₂ ---> 4P₂I₄ + 3P₄O₁₀

Question
Q: which reactant is in excess and by what mass?

Answer -

P₄O₆ moles = 0.02
I₂ moles  = 0.0118

------------------------------------------------------------

Now you can see that their stoichiometric coefficients are 5 and 8 for P₄O₆ and I₂ respectively. Therefore, for every 5 units of P₄O₆, 8 units of I₂ are used.

-------------------------------------------------------------

Now let us assume that P₄O₆ is the limiting reagent and is all used up.
So we will now calculate the actual moles of I₂ needed to completely react with 0.02 moles of P₄O₆ by the cross-multiplication method.
 
      5/0.02 = 8/x  (where x is mass of I₂ required)
=>   5x = 0.16
=>   x = 0.032 moles
 Since given I₂ is only 0.0118 moles, therefore our assumption contradicts.

--------------------------------------------------------------

This implies that P₄O₆ is not the limiting reagent but I₂ is the limiting reagent.

-----------------------------------------------------------------

 Now let us calculate moles of P₄O₆ needed to completely react with 0.0118 moles of I₂

      5/x = 8/0.0118
=>  8x = 0.059
=>   x = 0.007375 moles

Since less number of moles of P₄O₆ are required to completely react with 0.0118 moles of I₂ than the allotted 0.02 moles, therefore I2 is the limiting reagent and P₄O₆ is in excess.

---------------------------------------------------------------

Since only 0.007375 moles of P₄O₆ reacted, (0.02-0.007375 = 0.012625) moles remained unreacted.

---------------------------------------------------------------

Multiplying 0.012625 by 220 (molecular mass of P₄O₆), we get 2.7775g, which is the required answer.

Hope you understood it. Limiting reagents was my strong topic.

[Borek]

Answer -

Please read forum rules:

Don't give the final answers, let students solve questions on their own.

[Nisarg Shah]

Answer -

Please read forum rules:

Don't give the final answers, let students solve questions on their own.

Will keep that in mind