one question on this stiochemetry problem?

Question

ok now im really confused on how to do this one. Stiochemetry is very hard for me and i need to know how to do this question step by step please =) .....ok here it goes

Consider the following unbalanced equation
P4(s) + F2(g) -------> PF3(g)
How many grams of F2 are need to produce 120.g of PF3 if the reaction has 78.1% yield?

thanks guys! =)

Answer 1

Wow, I actually had a similar question on a test.
Anyway, first you need to figure out how much product you need to make such that 78.1% of it is exactly 120.g. You can do this by dividing 120.g by 78.1% which gives 154 g.
Next, you must always convert grams to moles, so ask yourself how many moles is 154g PF3? To figure that out, you divide 154g by the molar mass of PF3 which is P + F *3 = 31.0 + 19.0 * 3 = 88.0 g.
Dividing 154g by 88.0g equals 1.75 mols.
Now, notice that the equation is unbalanced, you always need to have a balanced equation: P4 + 6F2 -> 4PF3
Okay, now notice that for every 4 mols of PF3 you produce, you need 6 moles of PF3.
Since you need to produce 1.75 mols of PF3 you need:
1.75 mols * 6/4 = 2.63 mols F2
Now the final step is to conver mols F2 into grams, which you should be able to do.
The molar mass of F2 is 38.0. so 2.63 mols is 99.9 grams.
There, hopefully this helps!

Answer 2

As usual, you need to start with a balanced equation. Here, it would be P4 + 6F2 -> 4PF3. Calculate the molecular weights of PF3 and of F2, calculate the number of moles of PF3, find the number of moles of F2 from the equation, and multiply by the molecular weight of F2. Divide by the percent yield to get the amount of F2 needed.

Answer 3

P4(s) + 6F2(g) ===> 4PF3(g)

Atomic weights: P=31 F=19 F2=38 PF3=88

120.0gPF3 x 1molPF3/88gPF3 x 6molF2/4molPF3 x 38gF2/1molF2 x 78.1% = 60.7g F2

Answer 4

The first step whn you're given an unbalanced equation is to balance it. The balanced equation is P4 + 6F2 -------> 4PF3.

Next, we convert the desired amount of product from grams to moles. The molar mass of PF3 is 30.97 + 3(19) = 87.97 g/mol. so to find the number of moles needed, you divide the number of grams by the molar mass to get 1.482 moles of PF3.

Now, the next step is to work out how many moles of F2 would be needed if the reaction had 100% yield. Checking the balanced equation, we see that it takes 6 moles of F2 to make 4 moles of PF3. That means it takes 1 1/2 moles of F2 to make 1 mole of PF3. So to make 1.482 moles of PF3, it would take 2.223 moles of F2.

Now we consider the percent yield. Only 78.1% of the F2 used actually becomes PF3. So divide 2.223 by 0.781 to get 2.846 moles of F2. This is the amount you'd need to use.

Finally, you convert the amount of F2 from moles to grams. The molar mass of F2 is 19*2 = 38 g/mol, so you multiply the number of moles by the molar mass to get 108.164 grams of F2.