a chemist reacted 11.50 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 85%, what is the actual yield? show your work including, the use of stoichiometric calculations and conversion factors.

= 49.674 g NaCl

Explanation:

From the equation;

1 mole of Sodium metal produces 2 moles sodium chloride

This means;

23 g of Na will produce 116.88 g of NaCl

Therefore;

11.5 g will generate;

 = (11.5 × 116.88)/23

 =  58.44 g of NaCl

But;

Percentage yield = (Actual yield/Theoretical yield)× 100%

         85 /100 = Actual yield /58.44 g

Thus;

Actual yield = 0.85 × 58.44

                  = 49.674 g NaCl