Grams of F2 needed to produce 109 g of PF3 if 78.1% yield?

Question

P4 + 6F2 ---> 4PF3

What are the # grams of F2 are needed to produce 109 g of PF3 if the reaction has a 78.1% yield ?

Can anybody answer in sig figs: - I got 90.7 grams

Answer 1

I'm using F = 19.0 g/mol & P = 31.0 g/mol

Listed as 18.99840 & 30.97376 in CRC handbook 56th Ed.

[57.0/88.0] x 109 x 1/0.781 = 90.4g

The first part [57.0/88.0] is the fraction of how much of the PF3 is fluorine.

When multiplyed by 109g will give amount of fluorine needed if 100% yield.

The last term [1/0.781] corrects for only 78.1% yield.

My answer is slightly different (depends on what reference you get the molecular weights from, and how you round them off to 3 figures).

I think 3 significant figures is OK here, as the two given nunbers (109 & 78.1% are 3 significant figures).

Note: I don't mean to give the other guy a hard time, but my method is simpler, as you don't need to convert to moles (you don't even need to know the balanced equation; only molecular formula of the product).

Answer 2

I think there are 3 significant figures in this problem. Both 109 and 78.1 are 3 significant figures.
Okay, so you need 109g of product, but 109 is 78.1% yield. So that means that 109 = 0.781x. Solving for x gives you the mass of PF3 if you had 100% yield. x = 140g. The molecular weight of PF3 is 88.0 g/mol. Therefore you need 140/88.0 = 1.59 moles of PF3 if the yield were 100%. That means that you need 1.59 moles times 3 of F, since every mole of PF3 contains three Fluorines. Fluorine weighs 19.0 g/mol. But the tricky part is that the question asks for grams of F2, not just F; that's because fluorine gas is naturally in the form of F2, not F. So you really need 1.59 times 3/2 moles of F2. (Sorry if that's confusing. The product is F3 in nature, while the starting material is F2, so you need 3/2 the number of moles of F2.) 1.59 x 3/2 is 2.39 moles of F2. F2 weighs 38.0 g/mol. So you need 2.39 x 38.0 = 90.6 grams of F2.
We're one-tenth apart. So your answer is right, but go back and check to make sure you (or I) are always using the right significant figures. I could be the one making the mistake. But we got essentially the same answer, so we're right. Just possibly a difference in significant figures somewhere.
Good luck!

Answer 3

First write a balanced equation. Your equation isn't balanced. Calculate the molar mass of P and F from the periodic table. From the balanced equation, you will see which you want a million mol P and 3 mol F to produce a million mol of the compound. Calculate the molar mass of PF3 you may now say " X grams (calculated from moles X molar mass)F will produce Y grams (returned calculated from moles x molar mass) PF3. you like 122g PF3, so calculate via ratio and share. Yield is seventy 8.a million%, so multiply via one hundred/seventy 8.a million. i will say it is plenty greater complicated to describe the thank you to try this than it is to do the priority!! wish you stick to. If no longer email me and that i will do it for you with rationalization.