Given the following, what is the limiting reactant? What is the theoretical yield of #Cl_2#? If the yield of the reaction is 75.5%, what is the actual yield of chlorine?

Question

Chlorine gas can be prepared in the laboratory by the following reaction: #4HCl(aq) + MnO_2(s) -> MnCl_2(aq) + 2H_2O(l) + Cl_2(g)#. You add 40.5 g of #MnO_2# to a solution containing 41.7 g #HCl#.

Ellie Andrews · Accepted Answer

The actual yield is 3.83 g of #"Cl"_2"#.

This problem involves limiting reactants. We are aware that a balanced equation containing the masses, molar masses, and moles of the constituent compounds will be necessary. 1. Compile all the data in one location, placing the molar masses above the formulas and the rest below them. #M_r:color(white)(mmmm) 36.46 color(white)(ml)86.94color(white)(mmmmmmmmmmll)70.91# #color(white)(mmmmmm) "4HCl" + "MnO"_2 → "MnCl"_2 + "2H"_2"O" + "Cl"_2 # #"Mass/g:"color(white)(mml)41.7color(white)(mml)40.5# #"Amt/mol:"color(white)(ml)1.144color(white)(m)0.4658# #"Divide by:"color(white)(mm)4color(white)(mmmm)1# #"Moles rxn:"color(white)(ll)0.2859color(white)(m)0.4658# 2. Determine which reactant is limiting. Finding the "moles of reaction" that each will yield will make it simple to determine which reactant is the limiting one. In the balanced equation, you divide the moles of each reactant by the corresponding coefficient. In the table above, I completed that for you. #"HCl"# is the limiting reactant because it gives the fewest moles of reaction. 3. Calculate the theoretical moles of #"Cl"_2# #"Theoretical yield" = 0.2859 color(red)(cancel(color(black)("mol HCl"))) × ("1 mol Cl"_2)/(4 color(red)(cancel(color(black)("mol HCl")))) = "0.071 48 mol Cl"_2# 4. Calculate the theoretical yield of #"Cl"_2# #"Theoretical yield" = "0.071 48" color(red)(cancel(color(black)("mol Cl"_2))) × ("70.91 g Cl"_2)/(1 color(red)(cancel(color(black)("mol Cl"_2)))) = "5.069 g Cl"_2# 5. Determine the yield in actuality. #"Actual yield" = 5.069 color(red)(cancel(color(black)("g theoretical"))) × ("75.5 g actual")/(100 color(red)(cancel(color(black)("g theoretical")))) × 100 % = "3.83 g"# The actual yield of #"Cl"_2# is 3.83 g.

Owen Ambrose · Answer

undefined Need information on the reactants and their quantities to determine the limiting reactant and calculate theoretical yield and actual yield.