Help with these questions about limiting reactants and yield?
#1)# Given the reaction:
#3X + 4Y -> 5Z#
If after reacting #"2.1 mols"# of #X# with #"1.3 mols"# of #Y# , the percent yield is #50%# , and the actual yield is #"75.0 g"# of #Z# , what is the molar mass of #Z# ?
#2)# What is the theoretical yield of #"SiCl"_4# if #"2 g"# of #"Si"# react with #"1 g Cl"_2# ?
#"Si"(s) + 2"Cl"_2(g) -> "SiCl"_4(s)#
#3X + 4Y -> 5Z# If after reacting
#"2.1 mols"# of#X# with#"1.3 mols"# of#Y# , the percent yield is#50%# , and the actual yield is#"75.0 g"# of#Z# , what is the molar mass of#Z# ?
#"Si"(s) + 2"Cl"_2(g) -> "SiCl"_4(s)#
1 Answer
Note that these (mass + mols) must be based on theoretical yield, and not actual yield (because in actuality you may have lost reactant while the reaction went on, which reduces the true yield and creates errors in your calculations later on).
- You were given percent yield and actual yield, which allows you to find the theoretical mass yield.
- You were given the balanced chemical equation, which you can use to find the theoretical product quantity in
#"mol"# s.
So, we do the following calculations.
#"% Yield" = "Actual Yield"/"Theoretical Yield"#
#=> color(green)("Theoretical Yield") = "Actual Yield"/"% Yield"#
#= "75.0 g Z"/0.50 = color(green)("150.0 g Z")#
The
That means you have too little
Thus, your
#color(green)("mols Z") = "1.3 mols Y" xx "5 mols Z"/"4 mols Y" = color(green)("1.625 mols Z")#
(You should find that using
Therefore, your molar mass is:
#color(blue)(M_(m,Z)) = "150.0 g Z"/"1.625 mols Z"#
#=# #color(blue)("92.31 g/mol")#
In this case, try picking each reactant and converting to
That means assume one reactant is the limiting reactant, and check to see which one yields less product. In general:
#"? g SiCl"_4 = "? g A" xx "mols A"/"? g A" xx ("1 mol SiCl"_4)/"? mol A" xx ("? g SiCl"_4)/("mol SiCl"_4)#
#"? g SiCl"_4 = "? g B" xx "mols B"/"? g B" xx ("1 mol SiCl"_4)/"? mol B" xx ("? g SiCl"_4)/("mol SiCl"_4)#
Then, using the limiting reactant, you can find the theoretical yield of