In this reaction: #4HCl (g) + O_2 (g) -> 2H_2O (l) + 2Cl_2(g)#. When 63.1 g of #HCI# react with 17.2 g of #O_2#, 49.3 g of #Cl_2# are collected. How do you determine the limiting reactant, theoretical yield of #Cl_2# and percent yield for the reaction?

We must first identify the limiting reactant, and then we calculate the theoretical yield and percent yields.

We start with the balanced equation.

#color(white)(mmmmmmmm)"4HCl" + "O"_2 → "2H"_2"O" + "2Cl"_2#
#"MM/g·mol"^"-1":color(white)(ml) 36.46color(white)(m) 32.00color(white)(mmmmml) 70.91#

(a) Identify the limiting reactant

We calculate the amount of chlorine that can form from each reactant.

Calculate the moles of #"HCl"#

#"Moles of HCl" = 63.1 color(red)(cancel(color(black)("g HCl"))) × "1 mol HCl"/(36.46 color(red)(cancel(color(black)("g HCl")))) = "1.731 mol HCl"#

Calculate moles of #"Cl"_2# formed from the #"HCl"#

#"Moles of Cl"_2 = 1.731color(red)(cancel(color(black)("mol HCl"))) × "2 mol Cl"_2/(4 color(red)(cancel(color(black)("mol HCl")))) = "0.8653 mol Cl"_2#

Calculate the moles of #"O"_2#

#"Moles of O"_2 = 17.2 color(red)(cancel(color(black)("g O"_2))) × ("1 mol O"_2)/(32.00 color(red)(cancel(color(black)("g O"_2)))) = "0.5375 mol O"_2#

Calculate the moles of #"Cl"_2# formed from the #"O"_2#

#"Moles of Cl"_2 = 0.5375 color(red)(cancel(color(black)("mol O"_2))) × ("2 mol Cl"_2)/(1 color(red)(cancel(color(black)("mol O"_2)))) = "1.075 mol Cl"_2#

The limiting reactant is #"HCl"#, because it gives fewer moles of #"Cl"_2#.

(b) Calculate the theoretical yield of #"Cl"_2#.

#"Theoretical yield" = 0.8653 color(red)(cancel(color(black)("mol Cl"_2))) × "70.91 g Cl"_2/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "61.4 g Cl"_2#

The theoretical yield of #"Cl"_2# is 61.4 g.

(c) Calculate the percentage yield of #"Cl"_2#

The formula for percentage yield is

#color(blue)(|bar(ul(color(white)(a/a)"% yield" = "actual yield"/"theoretical yield" × 100 %color(white)(a/a)|)))" "#

#"% yield" = (49.3 color(red)(cancel(color(black)("g"))))/(61.4 color(red)(cancel(color(black)("g")))) × 100 % = 80.3 %#