Consider the following unbalanced reaction.

P4(s) + F2(g)---> PF3(g)
How many grams of F2 are needed to produce 114 g of PF3 if the reaction has a 80.3% yield?

I've worked it every way I can think, and the online grader says I am wrong. Someone plz lend me a hand.

well, there are about 50 ways to do it.

114 g of PF3 has how many grams of F in it? 114*atomicmassF/molmassPF3= 114(31/(50)=? you can work all that more accurately.

so how many grams of F2 is this? answer: 1/2 of the above.
With 80.3 percent yield? you need aboveF2 mass* (1/.803) grams F2

The conventional way to do this is with mole ratios, but it has to be balanced first.

To solve this problem, you need to follow a few steps:

Step 1: Write the balanced equation for the reaction:
P4(s) + 6F2(g) → 4PF3(g)

Step 2: Calculate the molar mass of PF3:
P: 1 * 30.97 g/mol = 30.97 g/mol
F: 3 * 18.99 g/mol = 56.97 g/mol
Total molar mass of PF3 = 30.97 g/mol + 56.97 g/mol = 87.94 g/mol

Step 3: Calculate the number of moles of PF3 produced:
114 g PF3 * (1 mol PF3 / 87.94 g PF3) = 1.295 mol PF3

Step 4: Use the stoichiometry of the balanced equation to find the number of moles of F2 required:
From the balanced equation, we know that 1 mol of PF3 is produced from 6 moles of F2. Therefore,
1.295 mol PF3 * (6 mol F2 / 4 mol PF3) = 1.9425 mol F2

Step 5: Calculate the mass of F2 needed:
1.9425 mol F2 * (38.00 g F2 / 1 mol F2) = 73.841 g F2

However, we know that the reaction has an 80.3% yield, so the actual amount of F2 needed would be higher. To calculate the actual amount, divide the calculated mass by the yield percentage:

73.841 g F2 / 0.803 = 91.9 g F2

Therefore, approximately 91.9 grams of F2 are needed to produce 114 g of PF3 with an 80.3% yield.

To solve this problem, you need to use stoichiometry and the given percent yield information. Let's break it down step by step:

Step 1: Write and balance the equation:
P4(s) + F2(g) ---> 4PF3(g)

Step 2: Determine the molar mass of PF3:
P: 30.97 g/mol
F: 18.99 g/mol
So, PF3: (30.97 g/mol) + (3 * 18.99 g/mol) = 87.94 g/mol

Step 3: Convert grams of PF3 to moles:
Given mass of PF3 = 114 g
Number of moles of PF3 = mass / molar mass = 114 g / 87.94 g/mol = 1.296 mol

Step 4: Determine the theoretical yield:
From the balanced equation, the stoichiometric ratio between F2 and PF3 is 1:4. So, for every 1 mole of F2, we get 4 moles of PF3.
Hence, moles of F2 = (moles of PF3) / 4 = 1.296 mol / 4 = 0.324 mol

Step 5: Convert moles of F2 to grams:
Molar mass of F2 = 2 * 18.99 g/mol = 37.98 g/mol
Mass of F2 = moles of F2 * molar mass of F2 = 0.324 mol * 37.98 g/mol ≈ 12.27 g

Step 6: Calculate the actual yield:
Given percent yield = 80.3%
Actual yield = percent yield * theoretical yield = 0.803 * 12.27 g ≈ 9.85 g

Therefore, you need approximately 12.27 grams of F2 to produce a theoretical yield of 114 grams of PF3. However, due to the 80.3% yield, the actual yield would be approximately 9.85 grams.