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 235 mL Cl2(g)235 mL Cl2(g) at 25 °C and 805 Torr805 Torr ?

Answer 1

Answer:

0.87g

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

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

Step 2:

Data obtained from the question. This includes the following:

Volume (V) of Cl2 obtained = 235mL

Temperature (T) = 25°C

Pressure (P) = 805 Torr

Step 3:

Conversion to appropriate unit.

For Volume:

1000mL = 1L

Therefore, 235mL = 235/1000 = 0.235L

For temperature:

Temperature (Kelvin) = temperature (celsius) + 273

Temperature (celsius) = 25°C

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

For Pressure:

760 Torr = 1 atm

Therefore, 805 Torr = 805/760 = 1.06 atm

Step 4:

Determination of the number of mole of Cl2 produced. This is illustrated below:

The number of mole (n) of Cl2 produced can be obtained by using the ideal gas equation as follow:

PV = nRT

Volume (V) = 0.235L

Temperature (T) = 298k

Pressure (P) = 1.06 atm

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

Number of mole (n)

PV = nRT

Divide both side by RT

n = PV /RT

n = (1.06 x 0.235)/(0.082 x 298)

n = 0.01 mole

Therefore 0.01 mole of Cl2 is produced from the reaction.

Step 5:

Determination of the number of mole MnO2 that produce 0.01 mole of Cl2. This is illustrated below:

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

From the balanced equation above,

1 mole of MnO2 produced 1 mole of Cl2.

Therefore, it will take 0.01 mole to MnO2 to also produce 0.01 mole of Cl2.

Step 6:

Converting 0.01 mole of MnO2 to grams.

This is illustrated below:

Number of mole MnO2 = 0.01 mole

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

Mass of MnO2 =?

Mass = number of mole x molar Mass

Mass of MnO2 = 0.01 x 87

Mass of MnO2 = 0.87g

Therefore, 0.87g of MnO2 is needed for the reaction.