Question

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 40.5 g MnO 2 is added to a solution containing 47.7 g HCl . What is the limiting reactant? HCl MnO 2 What is the theoretical yield of Cl 2 ? theoretical yield: 33.1 g Cl 2 If the yield of the reaction is 85.3 % , what is the actual yield of chlorine? actual yield: 28.23 g Cl 2

Answer 1

Answer:

1. HCl is the limiting reactant.

2. The theoretical yield of Cl2 is 23.197g

3. The Actual yield 19.787g

Explanation:

4HCl + MnO2 —> MnCl2 + 2H2O + Cl2

First let us calculate the number of mole of HCl and MnO2.

Molar Mass of HCl = 1 + 35.5 = 36.5g/mol

Mass of HCl = 47.7g

Number of mole = Mass /Molar Mass

Number of mole of HCl = 47.7/36.5 = 1.31mole

Molar Mass of MnO2 = 35 + (2x16) = 35 + 32 = 67g/mol

Mass of MnO2 = 40.5g

Number of mole = Mass /Molar Mass

Number of mole of MnO2 = 40.5/67 = 0.60mole

From equation,

4moles of HCl required 1mole of MnO2. Now let us consider the following:

4moles HCl require 1mole MnO2. Therefore 0.31mol of HCl will require = 0.31/4 = 0.0775mole of MnO2. This amount is little compared to the amount of MnO2 ( i.e 0.60mol) calculated. Therefore, HCl is the limiting reactant.

2. 4HCl + MnO2 —> MnCl2 + 2H2O + Cl2

Molar Mass of HCl = 36.5g/mol

Mass of HCl from the balanced equation = 4 x 36.5 = 146g

Molar Mass of Cl2 = 2 x 35.5 = 71g/mol

From the equation,

146g of HCl produced 71g of Cl2.

Therefore, 47.7g of HCl will produce = (47.7x 71)/146 = 23.197g of Cl2.

The theoretical yield of Cl2 is 23.197g

3. %yield = 85.3%

Theoretical yield = 23.197g

Actual yield =?

%yield = Actual yield /Theoretical yield

Actual yield = %yield x theoretical yield

Actual yield = 85.3% x 23.197g = (85.3/100) x 23.197 = 19.787g