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A mixture of 5.00 g of sodium carbonate, Na2CO3 and sodium hydrogen carbonate, NaHCO3 is heated. The loss in mass is 0.31 g. Sodium carbonate does not decompose on heating. Calculate the percentage by mass of sodium carbonate in the mixture.

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Anonymous

  1. I might argue with you that Na2CO3 does not decompose on heating but for this problem we will assume that is true.
    2NaHCO3 ==> Na2CO3 + H2O + CO2

    I don't see how to do with other than two simultaneous equations.
    Let X = g NaHCO3
    and Y = g Na2CO3
    -----------------
    eqn 1 is X + Y = 5.00

    Then you want to convert X to loss in mass due to CO2 and H2O.
    mm = molar mass

    (X/mm NaHCO3)*)1/2)*mmCO2 = g loss CO2
    (X/mm NaHCO3)*(1/2)*mmH2O = g loss H2O
    Add those together to get total loss which is 0.31g; i.e.,
    [(X/mm NaHCO3)*(1/2)*44] + [X/mm NaHCO3)*(1/2)*18 = 0.31 and solve this equation for X Plug that into equation 1 to find Na2CO3.
    Then %Na2CO3 = (g Na2CO2/5.00)*100 = ?

    Post your work if you get stuck.

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    DrBob222

  2. I saw the question and it’s answer its 83.2%

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    Anonymous

  4. Na2co3 doesn't decompose
    From the equation of reaction;
    168g of NaHco3-->6g of [co2+H2O]
    x------->0.31g
    x=0.84g
    Mass of Na2co3=5-0.84=4.16g

    %by mass of Na2co3=[4.16/5]×100
    =83.2%

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    Dr Flash

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