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12 March, 04:54

The reaction of hydrochloric acid with potassium permanganate is described above. 526.64 g of hydrogen chloride is reacted with 229.19 g of potassium permanganate. Assuming the reaction goes to completion, calculate the mass of each product produced. g of manganese (II) chloride g of water g of chlorine g of potassium chloride

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  1. 12 March, 07:52
    0
    Equation of the reaction:

    2KMnO4 (aq) + 16HCl (aq) - -> 2MnCl2 (aq) + 2KCl (aq) + 5Cl2 (g) + 8H2O (aq)

    To calculate the limiting reagent, we need to calculate the number of moles of the reactants:

    KMnO4:

    Molar mass = (39 + 55 + (16*4))

    = 158 g/mol

    Number of moles = mass/molar mass

    = 229.19/158

    = 1.4506 mol

    HCl:

    Molar mass = 1 + 35.5

    = 36.5 g/mol

    Number of moles = 526.64/36.5

    = 14.428 mol

    By stoichiometry, 2 moles of KMnO4 reacted with 16 moles of HCl

    The limiting reagent:

    14.428 moles of HCl * 2 moles of KMnO4/16moles of HCl

    = 1.8035 moles of KMnO4 is required to react with 14.428 moles of HCl

    But there's 1.4506 moles of KMnO4. Therefore, KMnO4 is the limiting reagent.

    Mass of the products:

    KCl:

    2 moles of KMnO4 will produce 2 moles of KCl

    Moles of KCl = 1 * 1.4506 mol

    = 1.4506 mol

    Molar mass = 39 + 35.5 = 74.5 g/mol

    Mass of KCl = 74.5 * 1.4506

    = 108.07 g

    MnCl2:

    2 moles of KMnO4 will produce 2 moles of MnCl2

    Number of moles of MnCl2 = 1 * 1.4506

    = 1.4506 mol

    Molar mass = 55 + (35.5*2)

    = 126 g/mol

    Mass of MnCl2 = 1.4506 * 126

    = 182.78 g

    Cl2:

    2 moles of KMnO4 will produce 5 moles of Cl2

    Number of moles of Cl2 = 5/2 * 1.4506

    = 3.6265 mol

    Molar mass of Cl2 = 35.5*2

    = 71 g/mol

    Mass of Cl2 = 71 * 3.6265

    = 257.4815 g

    H2O:

    2 moles of KMnO4 will produce 8 moles of H2O

    Number of moles of H2O = 8/2 * 1.4506

    = 5.80 mol

    Molar mass of H2O = (1*2) + 16

    = 18 g/mol

    Mass of H2O = 18*5.80

    = 104.44 g
  2. 12 March, 08:00
    0
    (1) mass of KCl = 108.025g

    (2) mass of MnCl2 = 182.7 g

    (3) mass of Cl2 = 257.37 g

    (4) mass of H20 (water) = 104.4g

    Explanation:

    we start with determining the limiting factor

    molar mass of KMnO4 = 158g/mol

    molar mass of HCl = 36.5g/mol

    hence,

    number of moles of KMnO4 = mass / molar mass

    number of moles of KMnO4 = 229.19/158 = 1.45moles

    number of moles of HCl = 526.64/36.5 = 14.43 moles

    we chose the lowest number of moles from the reactants as the limiting factor

    hence the limiting factor is KMnO4.

    calculating the mass of the products:

    for KCl

    2moles of KMnO4 reacts to give 2moles of KCl

    there for KCl contains 1.45moles

    (1) mass of KCl = number of moles of KCl x molar mass of KCl = 1.45 x 74.5 = 108.025g

    (2) from mole ratio 2 mole of KMnO4 gave 2 moles of MnCl2

    therefore 1.45moles KMnO4 will give 1.45 MnCl2

    mass of MnCl2 = 1.45moles x molar mass MnCl2 = 1.45 x 126 = 182.7 g

    (3) from mole ratio

    2moles KMnO4 gave 5moles Cl2

    1.45 moles will give (5 x1.45) / 2 moles of Cl2

    mass of Cl2 = number of moles of Cl2 x molar mass Cl2 = 3.625 x 71 = 257.37 g

    (4) from mole ratio

    2moles KMnO4 gave 8moles water (H2O)

    1.45 moles will give (1.45x 8) / 2 moles of H20

    mass of H2O produced = number of moles of H20 x the molar mass of H20 = 5.8 x 18 = 104.4g
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