A chemist reacted 17.25 grams of sodium metal with an excess amount of chlorine gas. The chemical reaction that occurred is shown. Na + Cl2 → NaCl. If the percentage yield of the reaction is 88%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.

Actual yield for the reaction is 38.5 g of NaCl

Explanation:

The balanced equation is this:

2 Na(s) + Cl₂ (g) → 2NaCl (s)

Ratio is 2:2. Let's determine the moles of sodium metal that has reacted.

17.25 g . 1 mol /23 gl = 0.75 moles

0.75 moles of Na were  used to produced 0.75 moles of NaCl, according to the equation.

Let's determine the mass of NaCl, that were produced.

0.75 mol . 58.45 g / 1 mol = 43.8 g

43.8 will be the theoretical yield (100%). Let's determine the 88%.

(Yield reaction / Theoretical yield) . 100 = 88

Yield reaction / 43.8 g = 88 / 100

Yield reaction = 0.88 . 43.8 g → 38.5  g