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8 August, 01:41

A 1.268 g sample of a metal carbonate, MCO3, was treated 100.00 mL of 0.1083 M H2SO4, yielding CO2 gas and an aqueous solution of the metal sulfate. The solution was boiled to remove all of the dissolved CO2 and then was titrated with 0.1241 M NaOH. A 71.02 mL volume of the NaOH solution was required to neutralize the excess H2SO4. a) Write the balanced chemical equation for this reaction. b) What is the identity of the metal? c) How many grams of CO2 gas were produced?

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  1. 8 August, 02:02
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    a) MCO3 + H2SO4 → MSO4 + CO2 + H2O

    2NaOH + H2SO4 → Na2SO4 + 2 H2O

    b) The metal is barium

    c) There is 0.2827 grams of CO2 produced

    Explanation:

    Step 1: Data given

    Mass of metal carbonate = 1.268 grams

    Volume of 0.1083 M H2SO4 solution = 100.00 mL

    The metal sulfate solution was boiled to remove all of the dissolved CO2 and then was titrated with 0.1241 M NaOH.

    A 71.02 mL volume of the NaOH solution was required to neutralize the excess H2SO4

    Step 2: The balanced equation:

    MCO3 + H2SO4 → MSO4 + CO2 + H2O

    2NaOH + H2SO4 → Na2SO4 + 2 H2O

    Step 3: Calculate moles of H2SO4

    Moles H2SO4 = molarity * volume

    Moles H2SO4 = 0.1083 M * 0.100 L

    Moles H2SO4 = 0.01083 moles

    Step 4: Calculate moles of NaOH

    Moles NaOH = 0.1241 M * 0.07102 L

    Moles NaOH = 0.008814 moles

    Step 5: Calculate limiting reactant

    For 1 mol of H2SO4, we need 2 moles of NaOH to produce 1 mol of Na2SO4 and 2 moles of H2O

    NaOH is the limiting reactant. It will completely be consumed (0.008814 moles).

    H2SO4 is in excess. There will be consumed 0.008814/2 = 0.004407 moles of H2SO4. There will remain 0.01083 - 0.004407 = 0.006423 moles of H2SO4

    Step 6: Calculate moles of MCO3

    There will react 0.006423 moles of H2SO4 with MCO3

    MCO3 + H2SO4 → H2O + CO2 + MSO4

    For 1 mole H2SO4, we need 1 mole of MCO3

    For 0.006423 moles of H2SO4, we need 0.006423 moles of MCO3, there will also be produced 0.006423 moles of CO2

    Step 7: Calculate molar mass of MCO3

    Molar mass MCO3 = mass MCO3 / moles MCO3

    Molar mass MCO3 = 1.268 grams / 0.006423 moles

    Molar mass MCO3 = 197.4 g/mol

    Step 8: Calculate molar mass of the metal

    Molar mass of C = 12 g/mol

    Molar mass of O = 16 g/mol

    Molar mass of metal = 197.4 - (12 + 3*16) = 137.4 g/mol

    The metal with molar mass of 137.4 g/mol is barium

    The metal carbonate is BaCO3

    Step 9: Calculate mass of CO2 produced

    Mass CO2 = moles CO2 * molar mass CO2

    Mass CO2 = 0.006423 * 44.01 g/mol

    Mass CO2 = 0.2827 grams
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