What is the theoretical yield and percent yield of $"AgCl"$ if $"102.6996 g CsCl"$ reacts with excess $"AgNO"_3"$? The actual yield of $"AgCl"$ is $"77.938858 g"$.

Md Ariful Islam

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Md Ariful Islam

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This is a double replacement (double displacement reaction).

Balanced Equation
$"CsCl(aq)"+"AgNO"_3("aq")$$rarr$$"CsNO"_3("aq")+"AgCl(s)"$

Molar Masses

We need the molar masses of $"CsCl"$ and $"AgCl"$.

$"CsCl:"$$"168.358452 g/mol"$

$"AgCl:"$$"143.3212 g/mol"$

Theoretical Yield of Silver Chloride

Determine the number of moles of $"CsCl"$ by dividing the given mass of $"CsCl"$ by its molar mass. Then determine the number of moles of $"AgCl"$ by multiplying times ratio between $"AgCl"$ and $"CsCl"$ so that $"AgCl"$ is in the numerator. Then determine the theoretical yield of $"AgCl"$ by multiplying times its molar mass.

$102.6996cancel"g CsCl"xx(1cancel"mol CsCl")/(168.358452cancel"g CsCl")xx(1cancel"mol AgCl")/(1cancel"mol CsCl")xx(143.3212"g AgCl")/(1cancel"mol AgCl")="87.42674 g AgCl"$

Percent Yield
$"percent yield"=("actual yield")/("theoretical yield")xx100%"$

$"percent yield"=(77.938858"g AgCl")/(87.42674"g AgCl")xx100%"$

$"percent yield"="89.1476 %"$

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