Year 11 Stoich Q?

[Andiio]

Could you please explain with explanations and working out? Teacher didn't really go over limiting/excess and textbook fails ==
Thanks!

4.4g of P4O6 and 3.00g of I2 are mixed and allowed to react according to the equation:

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

a) Which reactant is in excess and by what mass?

[pooshwaltzer]

n(P4O6) = 4.4g / (4*98 + 6*16)gmol-1 = 0.009mol
n(I2) = 3.0g / (2*126.9)gmol-1 = 0.0118mol

P4O6 = 0.009mol / 5 = 0.0018
I2 = 0.0118mol / 8 = 0.001475

P4O6 in excess; I2 limiting reagent.

[Andiio]

n(P4O6) = 4.4g / (4*98 + 6*16)gmol-1 = 0.009mol
n(I2) = 3.0g / (2*126.9)gmol-1 = 0.0118mol

P4O6 = 0.009mol / 5 = 0.0018
I2 = 0.0118mol / 8 = 0.001475

P4O6 in excess; I2 limiting reagent.

I worked out that P4O6 was in excess but i'm not sure as to whether I am correct with HOW MUCH it is in excess by and am not sure of how to calculate that; btw, isn't the Mm of P4 = 30.97*4? (Phosphorus)

[pooshwaltzer]

Molar mass (molecular weight) of P is 30.9738 g/mol

So yes, you're right. I dunno WTF I had gotten bloody 98 from. Going to bed now, maybe that'll fix prob.

[Andiio]

Molar mass (molecular weight) of P is 30.9738 g/mol

So yes, you're right. I dunno WTF I had gotten bloody 98 from. Going to bed now, maybe that'll fix prob.

Haha mmm..
Could you please tell me how to calculate how much phosphorus oxide is in excess by before going to sleep? XD

[Russ]

How many mol of the limiting reagent do you have? -> How many mol of the excess reactant will this consume?

It's in excess by however much you have left after that.

[pooshwaltzer]

Calculate how much product is produced by each reactant. 
NOTE:  It does not matter which product is chosen, but the same product must be used for both reactants so that the amounts can be compared.

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

Choosing P2I4(s) as the reference product...

n(P4O6) = 4.4g / 220gmol-1 = 0.02mol * 4[P2I4]/5[P4O6] = 0.016 * 569.6gmol-1 = 9.1136g of P2I4
n(I2) = 3.0g / (2*126.9)gmol-1 = 0.0118mol * 4[P2I4]/8[I2] = 0.0059 * 569.6gmol-1 = 3.36064g of P2I4
I2 < P4O6; I2 is our limiting reagent, P4O6 being the excess.

Next, to find the amount of excess reactant, we must calculate how much of the non-limiting reactant actually did react with the limiting reactant.
n(I2) = 3.0g / (2*126.9)gmol-1 = 0.0118mol * 5[P4O6] / 8[I2] = 0.007375 * 220gmol-1 = 1.6225g of P4O6

1.6225g is the amount of P4O6 that reacted, not what is left over.  To find the amount of excess P4O6 remaining, subtract the amount that reacted from the amount in the original sample.
4.4g - 1.6225g = 2.7775 of excess P4O6 remaining post reaction.


Choosing P4O10(s) as the reference product...also works...same deal but replace P2I4 in your calcs with P4O10 and adjust molar ratios to commensurate.


PS - Soz about the delayed reply; didn't notice your request until this morning. Luckily I'm not at work yet so was able to provide solution.