A chemist reacted 57.50 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 86%, what is the actual yield? Show your work, including the use of stoichiometric calculations and conversion factors.

The actual yield is 125.65 g.

Explanation:

The balanced equation is: 2 Na + Cl₂ → 2 NaCl.From the stichiometry of this balanced equation: 2.0 moles of Na reacts with 1.0 mole of chlorine gas to give 2.0 moles of NaCl.We need to calculate the number of moles of Na (n = mass/atomic mass), n = (57.50 g / 22.989 g/mole of Na) = 2.5 mole.

Using cross multiplication and from stichiometry:

2.0 moles of Na → 2.0 moles of NaCl

2.5 moles  of Na → 2.5 moles of NaCl

Now, we can calculate the mass of the produced NaCl (m = n x molar mass)m = (2,5 mole x 58.443 g/mole of NaCl) = 146.1 g.This is the yield if the percentage of yield is 100%, but the percentage yield of the reaction is 86%.The actual yield = (146.1 g) (86/100) = 125.65 g.