Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq), as described by the chemical equation

MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g)

How much MnO2(s) should be added to excess HCl(aq) to obtain 155 mL Cl2(g) at 25 °C and 725 Torr?

Answer 1

first calculate mole of Cl₂

Use ideal gas equation for calculation of mole of Cl₂

Ideal gas equation

PV = nRT             where, P = atm pressure= 725 torr = 0.953947 atm,

V = volume in Liter = 155 ml = 0.155 L

n = number of mole = ?

R = 0.08205L atm mol^-1 K^-1 =Proportionality constant = gas constant,

T = Temperature in K = 25⁰C = 273.15+ 25 = 298.15 K

We can write ideal gas equation

n = PV/RT

Substitute the value

n = (0.953947 X 0.155)/(0.08205 X 298.15) = 0.00604 mole

mole of Cl₂ obtain = 0.00604 mole

balanced reaction is

MnO_2(s) + 4HCl_(aq)   MnCl_2(aq) + 2H₂O_(l) + Cl_2(g)

According to reaction 1 mole MnO₂ produce 1 mole Cl₂ molar retio beween MnO₂ to Cl₂ is 1:1 therefore to obtain 0.00604 mole Cl₂ required MnO₂ = 0.00604 mole

molar mass of MnO₂ = 86.9368 g/mol

gm of compound = no. of mole X molar mass

gm of MnO₂ should added = 0.00604 X 86.9368 = 0.525 gm

gm of MnO₂ should added = 0.525 gm