Question

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)

Answer 1

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

Explanation:

We interrogate the following reaction:

`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"`, `"% yield"` `=` `"Moles of product"/"Moles of reactant"xx100%`.

We have `"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

See the explanation for the process and solutions.

Explanation:

Balanced Equation

`"AgNO"_3("aq") + "KI(aq)"` `rarr` `"KNO"_3("aq") + "AgI(s)"`

The overall process for answering this question is:

`color(red)("given mass AgNO"_3"` `rarr` `color(red)"mol AgNO"_3"` `rarr` `color(blue)"mol AgI"` `rarr` `color(blue)"mass AgI"`

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

`color(red)("given mass AgNO"_3:`
`8.30xx10^(-1)"g AgNO"_3"`

In order to determine mol `"AgNO"_3"`, its molar mass must be determined.

`"Molar Mass"` `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)":` `(1xx107.8682"g/mol Ag")+(1xx14.007"g/mol N")+(3xx15.999"g/mol O")=color(red)("169.8722 g AgNO"_3"`

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

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

Multiply the given mass of `"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 %`