12/2018 11.59 PM 4 76.2/100 -Prnt Calculator Periodic Table tion 27 of 35 Saplin

Question

12/2018 11.59 PM 4 76.2/100 -Prnt Calculator Periodic Table tion 27 of 35 Sapling Learning Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese/(IV) a You add 42.5 g of Mn02 to a solution containing 50.9 g of HCI (a) What is the limiting reactant? HCI (b) What is the theoretical yield of Ch2? Number g CI, (c) If the yield of the reaction is 85.9%, what is the actual yield of chlorine? Number Hint Previous Gve Up & Vev Soulion Check Answer Next Ed

Explanation / Answer

4HCl(aq) + MnO2(s) ------------> MnCl2(aq) + 2H2O(l) + Cl2(g)
a.
no of moles of MnO2 = W/G.M.Wt
                     = 42.5/86.94 = 0.488moles
no of moles of HCl = W/G.M.Wt
                    = 50.9/36.5 = 1.4 moles
1 mole of MnO2 react with 4 moles of HCl
0.488 moles of MnO2 react with = 4*0.488/1   = 1.952 moles of HCl
HCl is limiting reactant
b.
4 moles of HCl react with MnO2 to gives 1 moles of Cl2
1.4 moles of HCl react with MnO2 to gives = 1*1.4/4 = 0.35 moles of Cl2
mass of Cl2 = no of moles * gram molar mass
             = 0.35*71 = 24.85g of Cl2
Theoretical yield = 24.85g
c.
percent yield = actual yiled*100/Theoretical yiled
85.9           = 24.85*100/theoretical yiled
theoretical yiled = 24.85*100/85.9   = 28.93g of Cl2

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