What is the percent yield of PI3(g)?

Question

Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide:

     2 P(s) + 3 I₂(g) → 2 PI₃(g)

If 131 g of I₂(s) are reacted with excess P(s) to produce 0.306 mol of PI₃(g), what is the percent yield of PI₃(g)? 

Answer 1

Where to start:

Identify what you're looking for:

Percent yield

Identify formula for % yield

% yield = (actual / theoretical) x 100

Identify given and what we're missing

actual = 0.306 mol PI₃

theoretical = ?

Theoretical yield is calculated using the limiting reactant. The limiting reactant is the reactant that makes less product. Since it says "...reacted with excess P" the other reactant I₂ is your limiting reactant.

Convert 131 g I₂ to mol of PI₃

131g I₂ x (1 mole I₂/ 253.8g I₂) x (2 mols PI₃/3 mol I₂) = 0.34 mols PI₃

You can't convert directly from grams I₂ → mols PI₃

You have to use this pathway: grams I₂ → moles I₂ → moles PI₃

Use 1 mole/molar mass to convert from g to mols

Use the coefficients from the balanced equation to convert from moles I₂ → moles I₂

Theoretical yield:

0.34 mols PI₃

Plug into % yield formula:

(0.306/0.34) x 100 = 90% - ANSWER