Question

Chlorine can be prepared in the laboratory by the reaction of manganese dioxide with hydrochloric acid, HCl(aq)HCl(aq) , as described by the chemical equation MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) MnO2(s)+4HCl(aq)⟶MnCl2(aq)+2H2O(l)+Cl2(g) How much MnO2(s)MnO2(s) should be added to excess HCl(aq)HCl(aq) to obtain 295 mL Cl2(g)295 mL Cl2(g) at 25 °C and 795 Torr795 Torr ?

Answer 1

Answer:

1.131g of MnO2

Explanation:

Step 1:

The balanced equation for the reaction.

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

Step 2:

Data obtained from the question. This includes:

Volume (V) of Cl2 = 295mL = 0.295L

Temperature (T) = 25°C = 25°C + 273 = 298K

Pressure (P) = 795 Torr = 795/760 = 1.05 atm

Number of mole (n) of Cl2 =.?

Gas constant (R) = 0.082atm.L/Kmol

Step 3:

Determination of the number of mole of Cl2 produced. The number of mole of Cl2 produced from the reaction can be obtained by applying the ideal gas equation as follow:

PV = nRT

1.05 x 0.295 = n x 0.082 x 298

Divide both side by 0.082 x 298

n = (1.05 x 0.295)/(0.082 x 298)

n = 0.013 mole.

Therefore, the number of mole of Cl2 produced is 0.013 mole.

Step 4:

Determination of number of mole of MnO2 needed to produce 0.013 mole of Cl2.

This is illustrated below:

From the balanced equation above,

1 mole of MnO2 produced 1 mole of Cl2.

Therefore, it will also take 0.013 mole of MnO2 to produce 0.013 mole of Cl2.

Step 5:

Converting 0.013 mole of MnO2 to grams. This is illustrated below:

Number of mole MnO2 = 0.013 mole

Molar Mass of MnO2 = 55 + (2x16) = 87g/mol

Mass of MnO2 =?

Mass = number of mole x molar Mass

Mass of MnO2 = 0.013 x 87

Mass of MnO2 = 1.131g

Therefore 1.131g of MnO2 should be added to excess HCl.