Question

Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide:

4HCL(aq)+MnO2(s)------>MnCl2(aq)+2H2O(l)+Cl2(g)

You ad 39.3g of MnO2 to a solution containing 49.3g of HCl

a. what is the limiting reactant?

b. what is the theoretical yield of Cl2?

c. if the yield of the reaction is 84.1%, what is the actual yield of chlorine?

Answer 1

a)

Molar mass of HCl,

MM = 1*MM(H) + 1*MM(Cl)

= 1*1.008 + 1*35.45

= 36.458 g/mol

mass(HCl)= 49.3 g

use:

number of mol of HCl,

n = mass of HCl/molar mass of HCl

=(49.3 g)/(36.46 g/mol)

= 1.352 mol

Molar mass of MnO2,

MM = 1*MM(Mn) + 2*MM(O)

= 1*54.94 + 2*16.0

= 86.94 g/mol

mass(MnO2)= 39.3 g

use:

number of mol of MnO2,

n = mass of MnO2/molar mass of MnO2

=(39.3 g)/(86.94 g/mol)

= 0.452 mol

4 mol of HCl reacts with 1 mol of MnO2

for 1.352 mol of HCl, 0.3381 mol of MnO2 is required

But we have 0.452 mol of MnO2

so, HCl is limiting reagent

Answer: HCl is limiting reagent

b)

we will use HCl in further calculation

Molar mass of Cl2 = 70.9 g/mol

According to balanced equation

mol of Cl2 formed = (1/4)* moles of HCl

= (1/4)*1.352

= 0.3381 mol

use:

mass of Cl2 = number of mol * molar mass

= 0.3381*70.9

= 23.97 g

Answer: 24.0 g

c)

% yield = actual mass*100/theoretical mass

84.1= actual mass*100/23.97

actual mass=20.16 g

Answer: 20.2 g