Calculate the percentage of yield when 20 grams of sodium chloride solution reacts with an excess amount of silver nitrate solution knowing that 45 grams of silver chloride precipitated

Percentage % yield = 91.8%

Given:

20 g NaCl

45 g AgCl

To find:

% yield=?

The balanced chemical equation will be:

NaCl + AgNO₃ ⇒ AgCl + NaNO₃

First, we need to calculate moles of NaCl.

Then, calculate moles of AgCl from equation:

Mass AgCl(theoretical) :

Now, calculate for percentage yield.

The percentage yield will be 91.8%.

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%yield = 91.8

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

there all right good

explanation: