I posted this question earlier, but I believe it was a bit confusing in the way I typed it.
Given the following two step reaction:
4 P + 5 O2 -> P4O10
P4O10 + 6 H2O -> 4H3PO4
If you have 8.28 g of P, 15.65 g of O2, and 19.95 g of H2O, how many grams of H3PO4 can be produced?
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I've converted everything to moles:
4P:
8.28g
m=g/mw
x=8.28/124.0
x=0.067
502:
15.65g
x=15.65g/160
x= 0.0978
6H2O:
19.95g
x=19.96/108.0
0.1848
However, now I am stuck.
Do I need to find the limited?
If so, how do I go about this?
Please and Thank You
And it still is confusing. Yes, you do need to go through a limiting reagent. First, the molar mass of P (atomic mass) is 30.97 BUT P actually is P4 so let's re-write the equation as
P4 + 5O2 --> P4O10
moles P = 8.28/123.895 = 0.06683 mols P.
moles O = 15.65/32 = 0.4891 moles oxygen.
We have two scenarios depending upon which is the limiting reagent. I prefer to do it this way; i.e., determine the moles P4O10 produced under each.
a. If we have 0.06683 moles P4 and all the oxygen we need, how much P4O10 will we get?
0.06683 x (1 mole P4O10/1 mole P4) = 0.06683 x (1/1) = 0.06683 moles P4O10.
b. If we have 0.4891 moles oxygen and all the P4 we need.
0.4891 x (1 mole P4O10/5 moles O2) = 0.09781 moles P4O10.
c. The smaller number is ALWAYS the correct answer in limiting reagent problems; therefore, the limiting reagent is P4 and we have 0.06683 reacting and producing 0.06683 moles P4O10.
Now you are ready to tackle the second part of the reaction. It's another limiting reagent problem. Just follow the steps above.