Calculate the percentage of yield when 20 grams of sodium chloride solution reacts with an excess amount of silver nitrate solut ion knowing that 45 grams of silver chloride precipitated
1 answer:
%yield = 91.8
<h3>Further explanation</h3>
Given
20 g NaCl
45 g AgCl
Required
%yield
Solution
Reaction
NaCl + AgNO₃ ⇒ AgCl + NaNO₃
mol NaCl :
= mass : MW
= 20 g : 58,44 g/mol
= 0.342
mol AgCl from equation :
= 1/1 x mol NaCl
= 1/1 x 0.342
= 0.342
Mass AgCl(theoretical) :
= mol x MW
= 0.342 x 143,32 g/mol
= 49.02 g
%yield = (actual/theoretical) x 100%
%yield = (45/49.02) x 100%
%yield = 91.8
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