Consider the following unbalanced reaction.

P4(s) + F2(g) PF3(g)
What mass of F2 is needed to produce 212 g of PF3 if the reaction has a 84.2% yield?

Here is a worked example of a stoichiometry problem. http://www.jiskha.com/science/chemistry/stoichiometry.html

Since you want 212 grams AFTER an 84.2% yeild, you want to start with 212/0.842 = grams PF3.
Post your work if you get stuck.

To determine the mass of F2 needed, we first need to balance the chemical equation.

The unbalanced equation is:
P4(s) + F2(g) → PF3(g)

To balance this equation, we need to make sure that the number of atoms of each element is the same on both sides of the equation.

Balanced equation:
P4(s) + 6F2(g) → 4PF3(g)

Now we can calculate the molar mass of PF3.

Molar mass of PF3 = atomic mass of Phosphorus + 3 * atomic mass of Fluorine
= (31.0 g/mol) + 3 * (19.0 g/mol)
= 31.0 g/mol + 57.0 g/mol
= 88.0 g/mol

Next, we can calculate the number of moles of PF3 produced using the given mass and the molar mass.

Number of moles of PF3 = mass of PF3 / molar mass of PF3
= 212 g / 88.0 g/mol
= 2.41 mol

Since the reaction has an 84.2% yield, the actual amount of PF3 produced would be 84.2% of 2.41 mol.

Actual moles of PF3 produced = 84.2% * 2.41 mol
= 2.03 mol

From the balanced equation, we can see that 6 moles of F2 is required to produce 4 moles of PF3.

Based on this ratio, we can calculate the moles of F2 needed.

moles of F2 needed = (6/4) * moles of PF3
= (6/4) * 2.03 mol
= 3.04 mol

Finally, we can calculate the mass of F2 needed using the moles of F2 and its molar mass.

mass of F2 needed = moles of F2 * molar mass of F2
= 3.04 mol * 38.0 g/mol
= 115.52 g

Therefore, approximately 115.52 g of F2 is needed to produce 212 g of PF3 with an 84.2% yield.