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)MnO2(s) should be added to excess HCl(aq) to obtain 115 mL Cl2(g)at 25 °C and 735 Torr?

mass:

g MnO2

Answer 1

PV = nRT

where, P = pressure = 735 torr = 0.967 atm

V = volume = 115 mL = 0.115 L

n = number of moles

R = gas constant

T = temperature = 25 + 273 = 298 K

0.967 * 0.115 = n * 0.0821 * 298

0.111 = n * 24.5

n = 0.111 / 24.5 = 0.00453 mole

The number of moles of Cl2 = 0.00453 mole

From the balanced equation we can say that

1 mole of Cl2 requires 1 mole of MnO2 so

0.00453 mole of Cl2 will require

= 0.00453 mole of Cl2 *(1 mole of MnO2 / 1 mole of Cl2)

= 0.00453 mole of MnO2

mass of 1 mole of MnO2 = 86.9368 g

so the mass of 0.00453 mole of MnO2 = 0.394 g

Therefore, the mass of MnO2 required would be 0.394 g