# When (8.30x10^-1) g of silver nitrate is dissolved in water and then mixed with excess potassium iodide solution (6.0000x10^-1) g of precipitate form. What is the percent yield of the reaction? (Answer to 3 s.d. and in proper scientific notation)

Apr 13, 2017

Unknown on the given data...........

#### Explanation:

We interrogate the following reaction:

$A g N {O}_{3} \left(a q\right) + K I \left(a q\right) \rightarrow A g I \left(s\right) \downarrow + K N {O}_{3} \left(a q\right)$

Silver iodide, a canary yellow solid, may be treated as insoluble in aqueous solution, the losses probably occur on handling.........

For a $\text{1:1 rxn}$, $\text{% yield}$ $=$ "Moles of product"/"Moles of reactant"xx100%.

We have $\text{moles of reactant}$ (or at least we can calculate it), we have no data on the mass, i.e. the moles of product..........

Apr 13, 2017

See the explanation for the process and solutions.

#### Explanation:

Balanced Equation

$\text{AgNO"_3("aq") + "KI(aq)}$$\rightarrow$$\text{KNO"_3("aq") + "AgI(s)}$

The overall process for answering this question is:

color(red)("given mass AgNO"_3"$\rightarrow$$\textcolor{red}{\text{mol AgNO"_3}}$$\rightarrow$$\textcolor{b l u e}{\text{mol AgI}}$$\rightarrow$$\textcolor{b l u e}{\text{mass AgI}}$

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color(red)("given mass AgNO"_3:
$8.30 \times {10}^{- 1} \text{g AgNO"_3}$

In order to determine mol $\text{AgNO"_3}$, its molar mass must be determined.

$\text{Molar Mass}$ $M \text{(AgNO"_3)} :$

Multiply the subscript of each element by its atomic weight on the periodic table in g/mol, then add them.

$M \text{(AgNO"_3)} :$(1xx107.8682"g/mol Ag")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")=color(red)("169.8722 g AgNO"_3"

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color(red)("Moles AgNO"_3":

Multiply the given mass of $\text{AgNO"_3}$ by the reciprocal of its molar mass.

8.30xx10^-1color(red)cancel(color(black)("g AgNO"_3))xx(1"mol AgNO"_3)/(169.8722color(red)cancel(color(black)("g AgNO"_3)))=color(red)(4.8860xx10^(-3)"mol AgNO"_3"

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color(blue)("Moles AgI":

Multiply mol $\text{AgNO"_3}$ by the mol ratio $\left(1 {\text{mol AgI")/(1"mol AgNO}}_{3}\right)$ from the balanced equation.

4.8860xx10^(-3)color(blue)cancel(color(black)("mol AgNO"_3))xx(1"mol AgI")/(1color(blue)cancel(color(black)("mol AgNO"_3)))=color(blue)(4.8860xx10^(-3)"mol AgI"

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In order to determine the mass of $\text{AgI}$ produced from the given mass of $\text{AgNO"_3}$, the molar mass must be determined.

$M \text{(AgI)} :$

(1xx107.8682"g/mol Ag")+(1xx126.90447"g/mol I")=color(blue)("234.773 g/mol AgI"

color(blue)("Mass AgI":

Multiply mol $\text{AgI}$ by its molar mass.

$4.8860 \times {10}^{- 3} \text{mol AgI"xx(234.773"g"/"mol" "AgI")=color(blue)(1.15xx10^0"g AgI}$

The theoretical yield of $\text{AgI}$ from $8.30 \times {10}^{- 1} \text{g AgNO"_3}$ is $1.15 \times {10}^{0} \text{g AgI}$.

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"% yield"=("actual")/("theoretical")xx100

"% yield"=(6.0000xx10^(-1)"g AgI")/(1.15xx10^0"g AgI")xx100=5.22xx10^1 %