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)

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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)

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Answer 1 · 2017-04-13T09:04:52

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

Explanation:

We interrogate the following reaction:

$A g N O_{3} (a q) + K I (a q) \to A g I (s) ⏐ ↓ + K N O_{3} (a q)$ $AgNO_3(aq) + KI(aq) rarr AgI(s)darr + KNO_3(aq)$

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

For a $1:1 rxn$ $"1:1 rxn"$ , $% yield$ $"% yield"$ $=$ $=$ $\frac{Moles of product}{Moles of reactant} \times 100 %$ $"Moles of product"/"Moles of reactant"xx100%$ .

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

Answer 2 · 2017-04-13T20:17:44

See the explanation for the process and solutions.

Explanation:

Balanced Equation

${AgNO}_{3} (aq) + KI(aq)$ $"AgNO"_3("aq") + "KI(aq)"$ $\to$ $rarr$ ${KNO}_{3} (aq) + AgI(s)$ $"KNO"_3("aq") + "AgI(s)"$

The overall process for answering this question is:

${given mass AgNO}_{3}$ $color(red)("given mass AgNO"_3"$ $\to$ $rarr$ ${mol AgNO}_{3}$ $color(red)"mol AgNO"_3"$ $\to$ $rarr$ $mol AgI$ $color(blue)"mol AgI"$ $\to$ $rarr$ $mass AgI$ $color(blue)"mass AgI"$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

${given mass AgNO}_{3} :$ $color(red)("given mass AgNO"_3:$
$8.30 \times 10^{- 1} {g AgNO}_{3}$ $8.30xx10^(-1)"g AgNO"_3"$

In order to determine mol ${AgNO}_{3}$ $"AgNO"_3"$ , its molar mass must be determined.

$Molar Mass$ $"Molar Mass"$ $M {(AgNO}_{3}) :$ $M"(AgNO"_3)":$

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

$M {(AgNO}_{3}) :$ $M"(AgNO"_3)":$ $(1 \times 107.8682 g/mol Ag) + (1 \times 14.007 g/mol N) + (3 \times 15.999 g/mol O) = {169.8722 g AgNO}_{3}$ $(1xx107.8682"g/mol Ag")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")=color(red)("169.8722 g AgNO"_3"$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

${Moles AgNO}_{3} :$ $color(red)("Moles AgNO"_3":$

Multiply the given mass of ${AgNO}_{3}$ $"AgNO"_3"$ by the reciprocal of its molar mass.

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$color(blue)("Moles AgI":$

Multiply mol $"AgNO"_3"$ by the mol ratio $(1"mol AgI")/(1"mol AgNO"_3)$ from the balanced equation.

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

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

$M"(AgI)":$

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

$color(blue)("Mass AgI":$

Multiply mol $"AgI"$ by its molar mass.

$4.8860xx10^(-3)"mol AgI"xx(234.773"g"/"mol" "AgI")=color(blue)(1.15xx10^0"g AgI"$

The theoretical yield of $"AgI"$ from $8.30xx10^(-1)"g AgNO"_3"$ is $1.15xx10^0"g AgI"$ .

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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

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

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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)

2 Answers

Explanation:

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