Question

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

mass:

g MnO2

Homework Answers

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

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