[steph_r]

Hello all!

I have been at this question for so long now and still cant get the answer. Could i please have some help?

QUESTION:
4.40g of P₄0₆ and 3.00g of I₂ are mixed and allowed to react, according the equation: 

5P₄O₆(S) + 8I₂ (s)  4P₂I₄ (s) + 3P₄O₁₀ (s)

Which reactant is in excess and by what mass?

MY ATTEMPT:

n(P₄0₆) = 4.4/219.895 = 0.02001 mol
n(I₂) = 3.00/253.808 = 0.01182 mol

P₄O₆ : 0.02001/5 = 0.004                   (in excess)
I₂      : 0.01182/8 = 0.00148                (limiting reactant)

If P4O6 is in excess,
0.02001 - 0.004 = 0.016,
m(P4O6) = 0.016/219.88 = 3.52g is in excess

Actual answer is: 2.78g in excess

HELP WOULD BE GREATLY APPRECIATED :d

[Astrokel]

hello,

you cannot just subtract the amount like this because of the stoichiometric ratio. calculate the amount of P₄O₆ in excess by using the ratio first.

[steph_r]

Would you be able to show me step by step perhaps? Just that part?
Thank you for your patience

[Astrokel]

hey! i can't do entirely for you...so do you understand the working below?

amount of P4O6 reacted = (0.01182/8) x 5
amount of P4O6 in excess = ?
mass of P4O6 in excess = ?

[ARGOS++]

Dear steph_r;

Good start!,   -  Feed the following Diagram with your data and you will easily get the correct answer:

"Stoichiometry Problem (#23'025)"

Therein is also an Example Diagram:  How to do a "Stoichiometry Problem".

Good Luck!
                    ARGOS⁺⁺