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
