Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide.

4HCl(aq)+MnO2(s)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

A sample of 33.9 g MnO2 is added to a solution containing 47.1 g HCl.

What is the limiting reactant?

What is the theoretical yield of Cl2?

If the yield of the reaction is 79.3%, what is the actual yield of chlorine?

Answer 1

dus) given, mass of MnO2 = 33.99 mass of nel = 47.1g Steh - find males of Men oz and Hola males of MnO₂ = mass = 35.99 -0.39 mol molaemass 86.94g/mol. moles of Hcd = 4701g 429 mol 1029 mol 36 46 abral = Sten I find limiting reagento a) balanced Ryn. 4 nel cap) + Mane, - Mnrda (02) + 21,04612 (I) a) mole hee Coffrirent of Man (3 = males; -0,39 = 0.39 officient b) mole hee Lofficient of HCl = 1029 = 0.322 " 0.322 <0:39 aso. Limiting reagent - HCl v Step III find theoretical yeild of cle. applied POAC in this Ryn. 4 HCl(a) + MnO₂ (5) Molde (al) + 2H2O + Cola (0)

dus) given, mass of MnO2 = 33.99 mass of nel = 47.1g Steh - find males of Men oz and Hola males of MnO₂ = mass = 35.99 -0.39 mol molaemass 86.94g/mol. moles of Hcd = 4701g 429 mol 1029 mol 36 46 abral = Sten I find limiting reagento a) balanced Ryn. 4 nel cap) + Mane, - Mnrda (02) + 21,04612 (I) a) mole hee Coffrirent of Man (3 = males; -0,39 = 0.39 officient b) mole hee Lofficient of HCl = 1029 = 0.322 " 0.322 <0:39 aso. Limiting reagent - HCl v Step III find theoretical yeild of cle. applied POAC in this Ryn. 4 HCl(a) + MnO₂ (5) Molde (al) + 2H2O + Cola (0)

1. 29 moles of (da a moles of HCl cofficient officient mass of Cela molau mass of cle (70.1)x1 2 mads of Cely' = 1029X70.1 mass of Gole = 22064g Aco Theretical yeild of Celz = 220648 Step . hind actual yeåld of oz. actual yoild = Theoretical yeild xo hercoutage of it actual yild = 22.64 x 793 100 actual geild = 170959 Thank you, May it help you