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")##rarr##"CsNO"_3("aq")+"AgCl(s)"#

1 Answer
Meave60
Jan 13, 2016

The actual yield of AgCl is #"77.938858 g"#
The theoretical yield of AgCl is #"87.42674 g"#.
The percent yield of AgCl is #"89.1476 %"#.

Explanation:

This is a double replacement (double displacement reaction).

Balanced Equation
#"CsCl(aq)"+"AgNO"_3("aq")##rarr##"CsNO"_3("aq")+"AgCl(s)"#

Molar Masses

We need the molar masses of #"CsCl"# and #"AgCl"#.

#"CsCl:"##"168.358452 g/mol"#
https://pubchem.ncbi.nlm.nih.gov/compound/24293

#"AgCl:"##"143.3212 g/mol"#
https://pubchem.ncbi.nlm.nih.gov/compound/24561

Theoretical Yield of Silver Chloride

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

#102.6996cancel"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
#"percent yield"=("actual yield")/("theoretical yield")xx100%"#

#"percent yield"=(77.938858"g AgCl")/(87.42674"g AgCl")xx100%"#

#"percent yield"="89.1476 %"#

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