Question

1.A. How many grams of potassium permanganate should be used to prepare 34 L of the chlorine gas at 21.9 °C and 12.0 psi? (1 atm = 14.70 psi)

2KMnO4 + 16HCl --> 2MnCl2 + 5Cl2 +2KCl + 8H2O

B. What if the yield of the chlorine in this reaction is only 85.0 %? How many grams of potassium permanganate should be used?

Answer 1

Given reaction is

2KMnO₄ + 16HCl --> 2MnCl₂ + 5Cl₂ +2KCl + 8H₂O

From the above stiochiometric equation it is clear that 2 moles of KMnO4 produces 5 moles of Cl2 on reaction with excess HCl

Given that

Volume of Chlorine gas produced = 34 L

Temperature = 21.9°C = 294.9 K

Pressure = 12.0 psi = 12.0 psi *1 atm /14.7 psi

= 0.816 atm

According to ideal gas equation,

PV = nRT

Number of moles of Chlorine gas produced =

0.816 atm * 34 L/(0.0821 L-atm/(mol-K) * 294.9 K)

= 1.146 moles

Molar mass of KMnO4 = 158.03 g/mol

From stiochiometric equation

2*158.03 g of KMnO4 produces 5 moles of Cl2 on reaction with excess HCl

? g of KMnO4 produces 1.146 moles of Cl2 on reaction with excess HCl

Amount of KMnO4 required = 1.146 moles * 2*158.03 g/5 moles

= 72.44 g

b)

If Yield of Cl2 is 85.0% , Amount of KMnO4 required is same as that of if yiels is 100%.

We know that

% Yield = Actual Yield *100/Theoretical Yield

Actual Yield of Cl2 in terms of moles = % Yield * Theoretical Yield /100

= 85 * 1.146/100

= 0.9741 moles

If Yield is 85% , 0.9741 moles of Cl2 is formed.

Volume of Cl2 produced at 12 psi (0.816 atm) and 294.9 K is

= nRT/P

= 0.9741 moles * 0.0821 L-atm/(mol-K) *294.9 K/0.816 atm

= 28.9 L

Volume of Cl2 produced if Yield is 85 % = 28.9 L

Amount of KMnO4 required = 72.44 g to get 28.9 L of Cl2 at 0.816 atm and 294.9 K if the yield is 85%