#"13.1 g CaO"# react with #"38.9 g CO"_2"#. What is the percent yield if #"16.7 g CaCO"_3"# are produced?

1 Answer
Mar 16, 2017

Answer:

The limiting reactant is #"CaO"#. The percent yield is #"71.4%#.

Explanation:

Balanced equation.

#"CaO + CO"_2##rarr##"CaCO"_3"#

This is a limiting reactant question. We need to determine the mass of #"CaCO"_3"# that can be produced by the given masses of #"CaO"# and #"CO"_2"#. The one that produces the least mass of #"CaCO"_3"# is the limiting reactant, and the mass of #"CaCO"_3"# produced by that reactant is the theoretical yield. Once we have the mass of #"CaCO"_3"# produced, we can determine percent yield.

The process will include the following steps:

#"mass reactant"##rarr##"mol reactant"##rarr##"mol product"##rarr##"mass product"#

Since there are no coefficients, we know that all of the mole ratios between the compounds are 1:1.

We need the molar masses for all compounds.
#"CaO":##"56.077 g/mol"#
#"CO"_2":##"44.009 g/mol"#
#"CaCO"_3":##"100.086 g/mol"#
https://www.ncbi.nlm.nih.gov/pccompound?term=CaO
https://www.ncbi.nlm.nih.gov/pccompound/?term=CO2
https://www.ncbi.nlm.nih.gov/pccompound/?term=CaCO3

We need to determine the moles of each reactant by dividing the given masses by their molar masses.

#(13.1cancel("g") "CaO")/(56.077cancel"g"/"mol")="0.23361 mol CaO"#

#(38.9cancel("g") "CO"_2)/(44.009cancel("g")/"mol")="0.88391 mol CO"_2"#

Convert mol #"CaO"# to mol #"CaCO"_3"# to mass #"CaCO"_3"# by multiplying mol #"CaO"# by the mol ratio that gives mol #"CaO"# then multiply by the molar mass of #"CaCO"_3"#.

#0.23361cancel"mol CO"xx(1cancel"mol CaCO"_3)/(1cancel"mol CO")xx100.086"g CaO"_3="23.4 g CaO"_3"#

#color(blue)"13.1 g CaO"_3"##color(blue) "reacts with"# #color(blue) "CO"_2##color(white)(.)color(blue)"to produce"# #color(blue)("23.4 g CaO"_3"#.

#0.88391cancel("mol CO"_2)xx(1cancel"mol CaCO"_3)/(1cancel"mol CO"_2)xx(100.086"g CO"_2)/(1cancel"mol CaCO"_3)="88.5 g CaCO"_3"#

#color(blue)"38.9 g CO"_2"# #color(blue)"react with"# #color(blue)"CaO"##color(white)(.)color(blue)"to produce"# #color(blue)"88.5 g CaCO"_3"#.

The limiting reagent is #"CaO"#, so the maximum amount of #"CaO"_3"# that can be produced is #"24.4 g"#. This is the theoretical yield.

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

#"percent yield"=("16.7 g CaCO"_3)/("23.4 g CaCO"_3)xx100="71.4%"#