A 10.0-g sample of sodium chloride was placed in 10.0 g of water. If 3.85 g of Cl2 was obtained, what was the percent yield of Cl2

63.53% yield

Explanation:

The balanced equation for this reaction is 2NaCl + H2O -> 2NaOH +Cl2

First we must find the limiting reactant

From NaCl we can only produce 6.06 grams of Cl2 in theory

From H20 we can only produce 38.995 grams in theory

so we know NaCl is the limiting

% yield is (Actual/Theoretical) x100 so

(3.85/6.06)x100= 63.53% yield

i'm pretty sure its b i might be wrong though!