53.0 g of silver (1) nitrate reacts according to the reaction shown below: AgNO3 + KCI — AGCI + KNO3
Find the theoretical yield of AgCl. If the amount of AgCl actually produced is 38.5 grams,
what is the percent yield of the experiment?

Percent yield = 86.09%

Theoretical yield of AgCl = 44.72 g

Explanation:

Given data:

Mass of silver(1) nitrate = 53.0 g

Actual yield of AgCl = 38.5 g

Theoretical yield of AgCl = ?

Percent yield = ?

Solution:

Chemical equation:

AgNO₃ + KCl    →   AgCl + KNO₃

Number of moles of AgNO₃:

Number of moles = mass/molar mass

Number of moles = 53.0 g/ 169.87 g/mol

Number of moles = 0.312 mol

now we will compare the moles of AgNO₃ and AgCl

               AgNO₃         :           AgCl

                   1               :                1

                0.312          :           0.312

Theoretical yield of AgCl:

Mass = number of moles × molar mass

Mass = 0.312 mol × 143.32 g/mol

Mass = 44.72 g

Percent yield:

Percent yield = (actual yield / theoretical yield) × 100

Percent yield = ( 38.5 g / 44.72 g ) × 100

Percent yield = 86.09%

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