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Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid,...

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?

Solutions

Expert Solution

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

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


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