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 MnO 2 ( s ) + 4 HCl ( aq ) ⟶ MnCl 2 ( aq ) + 2 H 2 O ( l ) + Cl 2 ( g ) How much MnO 2 ( s ) should be added to excess HCl ( aq ) to obtain 295 mL Cl 2 ( g ) at 25 °C and 775 Torr ?

Mass of MnO2:____g

Answer 1

1.069 g

Explanation

Number of moles of Cl2 required is calculated by ideal gas equation

PV = nRT

P = Pressure , 775torr = (1atm/760torr)× 775torr = 1.020atm

V = volume , 295ml = 0.295L

n = number of moles , ?

R = gas constant , 0.082057( L atm/ mol K)

T = Temperature , 25℃ = 298.15K

n = PV/RT

= 1.020atm × 0.295L/(0.082057(L atm/mol K) × 298.15K)

= 0.012299mol

moles of Cl2 required to produce = 0.012299mol

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

stoichiometrically , to produce 1mole of Cl2 1mole of MnO2 required

moles of MnO2 required to produce 0.012299moles of Cl2 = 0.012299mol

mass = number of moles × molar mass

mass of MnO2 should be added = 0.012299mol × 86.94 g/mol = 1.069 g