What is the limiting reactant? What is the theoretical yield of Cl2?
Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl ( aq ) + MnO 2 ( s ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g ) A sample of 40.1 g MnO 2 is added to a solution containing 43.7 g HCl
- Roger the MoleLv 71 month agoFavourite answer
4 HCl + MnO2 → MnCl2 + 2 H2O + Cl2
(40.1 g MnO2) / (86.93685 g MnO2/mol) = 0.461254 mol MnO2
(43.7 g HCl) / (36.4611 g HCl/mol) = 1.19854 mol HCl
1.19854 moles of HCl would react completely with 1.19854 x (1/4) = 0.299635 mole of MnO2, but there is more MnO2 present than that, so MnO2 is in excess and HCl is the limiting reactant.
(1.19854 mol HCl) x (1 mol Cl2 / 4 mol HCl) x (70.9064 g Cl2/mol) =
21.2 g Cl2 in theory