Question

Attempt 3 < Question 11 of 12 > Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide, 4 HCl(aq) + MnO,(8) -MnCl(aq) + 2 H2O(1) + Cl (8) A sample of 38.1 g Mno, is added to a solution containing 42.5 g HCI. What is the limiting reactant? OHCI O Mno, O MNO2 What is the theoretical yield of Cl,? theoretical yield: 8 C, If the yield of the reaction is 79.3%, what is the actual yield of chlorine? actual yield:

Answer 1

Balanced equation:
MnO₂ + 4 HCl ====> MnCl₂ + 2 H₂O + Cl₂

Mass of MnO2 = 38.1 gm

Moles of MnO2 = 38.1 gm / 86.93 g/mol =  0.43824 Moles

Mass of HCl = 42.5 gm

Moles of HCl = 42.5 gm / 36.46 g/mol =  1.1656 Moles

Limiting reagent is HCl.

Moles of Chlorine is produced =  0.2914 Moles

Theoretical yield of Chlorine = 0.2914 Moles x 70.906 gm/mol =  20.662 gm

Yield of the reaction = 79.3 %

Yield of chlorine for 79.3 % = 20.662 gm x 79.3 % / 100 % = 16.38 gm .

Balanced equation: MnO2 + 4 HCI = MnCl2 + 2 H20+ Cl2 Reaction stoichiometry Limiting reagent Compound Coefficient Molar Mass Moles Weight MnO2 86.936845 0.2914078 25.33407 HCI 36.46094 1.165631 42.50001 MnCl2 125.844045 0.2914078 36.67193 HO 18.01528 0.5828156 10.49959 Cl2 70.906 0.2914078 20.66256 Units: molar mass - g/mol, weight - g.

< Question 11 of 12 > Attempt 3 Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl(aq) + Mn0,8) MnCl(aq) + 2H2O(1) + Cl2(E) A sample of 38.1 g Mno, is added to a solution containing 42.5 g HCI. What is the limiting reactant? OHCI Mno, What is the theoretical yield of Cl,? theoretical yield: 20.662 If the yield of the reaction is 79.3%, what is the actual yield of chlorine? actual yield: 16.38