Question

Question 8 of 11> Chlorine gas can be prepared in the laboratory by the reaction of hydrochloric acid with manganese(IV) oxide. 4 HCl(aq)+ MnO2(s) MnCl, (aq) +2 H,O0) + Cl,(g) A sample of 34.7 g MnO, is added to a solution containing 42.1g HCl. What is the limiting reactant? MnO2 НС What is the theoretical yield of Cl,? theoretical yield: If the yield of the reaction is 81.7 % , what is the actual yield of chlorine? g CI, actual yield: contact us help prvacy policy terms of use about us careers MacBook Pro

Answer 1

Molar mass of HCl = 1.0079 + 35.45 = 36.46 g/mol

Molar mass of MnO ₂ = 54.94 + 2(16.00) = 86.94 g/mol

Molar mass of Cl ₂ = 2(35.45) = 70.90 g /mol

Consider reaction, 4 HCl + MnO ₂$\rightarrow$ MnCl ₂ + 2 H₂O + Cl ₂

From above reaction, Stoichiometric ratio of HCl : MnO ₂ is 4:1.

Calculation of no. of moles

We have, No. of moles = Mass / Molar mass

No. of moles of HCl = 42.1 g / 36.46 g/mol = 1.155 mol

No. of moles of MnO ₂ = 34.7  g / 86.94 g/mol = 0.3991 mol

Therefore, provided molar ratio of HCl : MnO ₂ is 1.155 : 0.3991 = 2.894 :1

Comparing provided molar ratio with stoichiometric ratio , it is found that HCl is a limiting reactant.

Hence, yield of product will depend on amount of HCl.

Calculation of theoretical yield of chlorine.

From reaction, 4 mol HCl $\equiv$ 1 mol Cl ₂

i e 4 $\times$ 36.46 g HCl $\equiv$ 70.90 g Cl ₂

$\therefore$ 42.1 g HCl $\equiv$ 70.90 $\times$ 42.1 / 4 $\times$ 36.46 g Cl ₂

42.1 g HCl $\equiv$ 20.5 g Cl ₂

Theoretical yield of chlorine = 20.5 g

Calculation of Actual yield of chlorine.

We have formula, % yield = [ Actual yield / Theoretical yield ] 100

81.7 = [ Actual yield / 20.5 g ] 100

[ Actual yield / 20.5 g ] = 81.7 / (100) = 0.817

Actual yield = 0.817 ( 20.5 g) = 16.7 g

ANSWER : Actual yield of chlorine = 16.7 g