Very Difficultg Chemistry Question?

Question

Consider the following unbalanced reaction:

P4(s) + F2(g) --------> PF3(g)

How many grams of F2 are needed to produce 120. g of PF3 if the reaction has a 78.1% yield

Answer 1

P4 + 6 F2 -------  4 PF3

If 120 grams is produced as a result of a 78.1 % yield from theoretical..then the theoretical yield is 120/0.781 = 153.65 grams. 153 .65 grams of PF3 is 153.65/88grams/mol = 1.746 moles of PF3. Since 6 moles of F2 yield 4 moles of PF3 then

6/4 = X /1.746moles  = 2.62 moles of F2 2.62 moles X 38grams/mole = 99.56 grams of F2.

Proof

6 ( 38) grams of F2 will yield 4(88) grams of PF3

soo 228 grams/352 grams = Xgrams/153.65 grams

Xgrams = 99.54 grams

Answer 2

well, since the yield is 78.1%, you have to find what you need as your "theorhetical" value. so:

781/1000=120/x

cross multiply

120000=781x

I'll leave you to figure that out, but let's say it is around 150g. so we take that value into demnsional analysis;first being from grams to moles

150g PF3/(??g PF3=1 mol PF3)

the part under paranetheses is a ratio

so, you find the ratio by adding the molecular weights together

P=31

F=19*3

31+19*3=88

so

150g PF3/88g PF3=?? mol PF3

now our number of moles in the equation correspond: (after balancing)

P4 +F2--->PF3====> P4+6F2-->4PF3

so four moles of PF3 need six moles of F2

??mol PF3/(4 mol PF3/6mol F2)=?? mol F2

?? mol F2/38g F2= g F2

just explaining seems a little complicated, so in demensional analysis, your end result will look like this

150g PF3/1 mol PF3/ 6 mol F2/ 38g F2

--------------------------------------...

                 /88 g PF3 / 4 mol PF3/ 1 mol F2

everything on the top is multiplied and then divided by everything on the bottom. so:

150*1*6*38/(88*4*1)=  about 97

so around 97 grams of F2 are needed to make around 120g of PF3. now that you know the method, the complete work is up to you