# Calculate the percentage yield of ferrous sulfide if 3 moles of iron produces 220 grams of ferrous sulfide?

May 18, 2017

First we calculate the theoretic yield.

#### Explanation:

3 mol of $F e$ should yield 3 mol $F e S$

$m \left(F e S\right) = m \left(F e\right) + m \left(S\right) = 55.85 + 32.07 = 87.92$

3 mol should have a mass of $3 \times 87.92 = 263.76$

So percent yield is 220/263.76=0.834~~83%