#"364.0 g Sb"_2"S"_3"# reacts with excess #"Fe"# to produce #"35.0 g Sb"#. What is the percent yield?

#"Sb"_2"S"_3 + "3Fe"##rarr##"2Sb + 3FeS"#

1 Answer
Mar 14, 2017

Answer:

The theoretical yield is #13.4%#.

Explanation:

#"Sb"_2"S"_3" + 3Fe"##rarr##"2Sb + 3FeS"#

Determine the mole ratios between #"Sb"_2"S"_3# and #"2Sb"#.

#(1"mol Sb"_2"S"_3)/(2"mol Sb")# and #(2"mol Sb")/(1"mol Sb"_2"S"_3")#

The mass of #"Sb"_2"O"_3"# needs to be converted to moles. Divide the given mass by its molar mass, #"339.7 g/mol"#. https://pubchem.ncbi.nlm.nih.gov/compound/16685273

#(364.0color(red)(cancel(color(black)("g"))) "Sb"_2"S"_3)/(339.7color(red)cancel(color(black)("g"))/"mol")="1.0715 mol Sb"_2"S"_3"#

Multiply the moles #"Sb"_2"S"_3# by the mole ratio determined above with #"2 mol Sb"# in the numerator.

#1.0715color(red)(cancel(color(black)("mol"))) "Sb"_2"S"_3xx(2 "mol Sb")/(1color(red)(cancel(color(black)("mol") ))"Sb"_2"S"_3)="2.143 mol Sb"#

Multiply the moles #"Sb"# by its molar mass to determine the theoretical mass of #"Sb"#. Molar mass is the atomic weight on the periodic table in grams/mole, and for Sb it is #"121.76 g/mol"#.

#2.143 color(red)(cancel(color(black)("mol Sb")))xx(121.76"g Sb")/(color(red)(cancel(color(black)("mol Sb"))))="260.9 g Sb"#

Theoretical yield:

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

#"% yield"=(35.0 color(red)(cancel(color(black)("g"))))/(260.9 color(red)(cancel(color(black)("g"))))xx100="13.4 %"#