Answer to Question #324172 in Physical Chemistry for NOEMI

Question

100 g of manganese dioxide reacts with 300 g of hydrochloric acid in the reaction shown below.

MnO2 + 4HCl -> MnCl2+ 2H2O + Cl2 

 (Use the following molar masses: MnO2= 86.9g/mol HCl = 36.5g/mol ; MnCl2= 125.8g/mol; H2O= 18.02g/mol and Cl 2 =70.9 g/mol.

 a. Which reactant is limiting?

b. Which reactant is excess? 

c. What is the theoretical yield of Cl2? 

d. What is the percent yield of the reaction if 75.0 g of Cl2 was produced

Answer 1

a.      Which reactant is limiting?

Moles MnO₂:

100 / 86.9 = 1.15 (mol)

Moles HCl:

300 / 36.5 = 8.22 (mol)

The mole ratio of the reactants is 1 : 4. So, in order for 1.15 mol of MnO₂ to reacts 1.15*4 = 4.6 mol of HCl required. Thus, MnO₂ is limiting reagent.

 

b.      Which reactant is excess? 

HCl needed for the reaction is 4.6 mol. We’re given 8.22. So, HCl is in excess.

 

c.      What is the theoretical yield of Cl₂? 

The ratio between MnO₂ and Cl₂ is 1 : 1. So, moles of Cl₂ produced is 1.15 mol – theoretical yield.

Mass = 1.15 * 70.9 = 81.535 g.

 

d.      What is the percent yield of the reaction if 75.0 g of Cl2 was produced?

Percent yield = 75.0 / 81.535 = 0.92 or 92%.