Consider the following unbalanced reaction.
P4(s) + F2(g) PF3(g)
How many grams of F2 are needed to produce 212 g of PF3 if the reaction has a 79.8% yield?
I did 212/.798=265.66 grams
do I convert that to moles next?
i got 88 mol. is that right?
No. mols = grams/molar mass
265.66g x (1mol/88) = 3.01 mol PF3
very good. Now convert mols PF3 to mols F2 BUT make sure the equation is balanced first. I didn't check it--it may be balanced now.
the balanced equation is P4+6F2-->4PF3
so i have:
3.01mol PF3 x (6 molF2/4molPF3)=4.5mol2F2
So then i convert it to grams, correct?
I would have rounded the 265.66/88 = 3.0189 to 3.02, then 3.02 x 6/4 = 4.528 which I would round to 4.53 mols F2 and convert that to grams. Actually, I never round step by step. I leave those numbers in my calculator and use the answer to one as the first number in the next step. Something like this.
212/88 = ?
? x 6/4 = ?
? x molar mass F2 = ? grams, then round that to three significant figures since the 212 had three in the problem.
so is the answer 171 g F2?
4.5 mol F2 x (38 g/1mol) =171g
Thank you very much!
Almost. If you use the rest of the 4.5 (4.528) x 38 I get 172 g. Also, I see in my last response that I omitted a step but I think you have the process down ok.
How would you find the remaining mass of the excess reagent?posted by Brian