Question

Question 29 of 38 > Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl(aq) + MnO,(s) MnCl, (aq) + 2H,0() + CI,(8) A sample of 42.3 g Mno, is added to a solution containing 45.5 g HCL What is the limiting reactant? O Mno, OHCI What is the theoretical yield of CI? theoretical yield: If the yield of the reaction is 76.9%, what is the actual yield of chlorine? actual yield: Joch

Answer 1

1)

Molar mass of HCl,
MM = 1*MM(H) + 1*MM(Cl)
= 1*1.008 + 1*35.45
= 36.458 g/mol


mass(HCl)= 45.5 g

use:
number of mol of HCl,
n = mass of HCl/molar mass of HCl
=(45.5 g)/(36.46 g/mol)
= 1.248 mol

Molar mass of MnO2,
MM = 1*MM(Mn) + 2*MM(O)
= 1*54.94 + 2*16.0
= 86.94 g/mol


mass(MnO2)= 42.3 g

use:
number of mol of MnO2,
n = mass of MnO2/molar mass of MnO2
=(42.3 g)/(86.94 g/mol)
= 0.4865 mol
Balanced chemical equation is:
4 HCl + MnO2 ---> Cl2


4 mol of HCl reacts with 1 mol of MnO2
for 1.248 mol of HCl, 0.312 mol of MnO2 is required
But we have 0.4865 mol of MnO2

so, HCl is limiting reagent
Answer: HCl

2)
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)*1.248
= 0.312 mol


use:
mass of Cl2 = number of mol * molar mass
= 0.312*70.9
= 22.12 g
Answer: 22.1 g

3)
% yield = actual mass*100/theoretical mass
76.9= actual mass*100/22.12
actual mass=17.01 g
Answer: 17.0 g