ap chemistry

posted by Kris

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?

  1. DrBob222

    Yes.

  2. Kris

    i got 88 mol. is that right?

  3. DrBob222

    No. mols = grams/molar mass

  4. Kris

    265.66g x (1mol/88) = 3.01 mol PF3

    correct?

  5. DrBob222

    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.

  6. Kris

    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?

  7. DrBob222

    Yes, almost.
    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.

  8. Kris

    so is the answer 171 g F2?

    I did:
    4.5 mol F2 x (38 g/1mol) =171g

  9. Kris

    Thank you very much!

  10. DrBob222

    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.

  11. Brian

    How would you find the remaining mass of the excess reagent?

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