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10 February, 16:34

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?

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  1. 10 February, 20:06
    0
    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.
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