A chemist reacted 17.25 grams of sodium metal with excess chlorine gas. The chemical reaction is as follows

Na + Cl2 → NaCl

If the percentage yield of the reaction is 88%, what is the actual yield?

38.588 g ≅ 35.60 g.

Explanation:

From the balanced equation: 2Na + Cl₂ → 2NaCl, It is clear that 2.0 mole of Na reacts with 1.0 mole of Cl₂ to produce 2.0 mole NaCl. The no. of moles of Na metal reacted = mass / atomic mass = (17.25 g) / (22.989 g/mol) = 0.75 mol.

Using cross multiplication:

2.0 moles of Na metal produce → 2.0 moles of NaCl,

∴ 0.75 mole of Na metal produce → 0.75 mole of NaCl.

The calculated (theoretical) yield of NaCl = n x molar mass = (0.75 mol) (58.44 g/mol) = 43.85 g.

∵ The percentage yield = [ (actual yield) / (calculated yield) ] x 100.

∴ The actual yield = [ (the percentage yield) (calculated yield) ] / 100 = [ (88 %) (43.85 g) ] / 100 = 38.588 g ≅ 35.60 g.