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
Apr 13, 2017

Answer:

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..........

Apr 13, 2017

Answer:

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 %#