Question

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​

Answer 1

%yield = 91.8

Further explanation

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

Answer 2

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.

$\text{Number of Moles}=\frac{\text{Given Mass}}{\text{Molar Mass}} \\\\\text{Number of Moles}=\frac{20}{58.44} \\\\\text{Number of Moles}= 0.342 moles$

Then, calculate moles of AgCl from equation:

$\text{mol of AgCl}= 1/1 * \text{ mol NaCl}\\\\\text{mol of AgCl}= 1/1 * 0.342\\\\\text{mol of AgCl}= 0.342$

Mass AgCl(theoretical) :

$\text{Mass of AgCl}= mol * MW\\\\\text{Mass of AgCl}= 0.342 x 143,32 g/mol\\\\\text{Mass of AgCl}= 49.02 g$

Now, calculate for percentage yield.

$\%yield =\frac{actual}{theoretical} * 100\%\\\\\%yield = (45/49.02) * 100\%\\\\\%yield = 91.8\%$

The percentage yield will be 91.8%.

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