# 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".

## "CsCl(aq)"+"AgNO"_3("aq")$\rightarrow$$\text{CsNO"_3("aq")+"AgCl(s)}$

Jan 13, 2016

The actual yield of AgCl is $\text{77.938858 g}$
The theoretical yield of AgCl is $\text{87.42674 g}$.
The percent yield of AgCl is $\text{89.1476 %}$.

#### Explanation:

This is a double replacement (double displacement reaction).

Balanced Equation
"CsCl(aq)"+"AgNO"_3("aq")$\rightarrow$$\text{CsNO"_3("aq")+"AgCl(s)}$

Molar Masses

We need the molar masses of $\text{CsCl}$ and $\text{AgCl}$.

$\text{CsCl:}$$\text{168.358452 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/24293

$\text{AgCl:}$$\text{143.3212 g/mol}$
https://pubchem.ncbi.nlm.nih.gov/compound/24561

Theoretical Yield of Silver Chloride

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

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

Percent Yield
$\text{percent yield"=("actual yield")/("theoretical yield")xx100%}$

$\text{percent yield"=(77.938858"g AgCl")/(87.42674"g AgCl")xx100%}$

$\text{percent yield"="89.1476 %}$