Chemistry

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Stomach acid is essentially 0.10 M HCl. An active ingredient found in a number of popular antacids is CaCO3. Calculate the number of CaCO3 needed to exactly react with 250 mL of stomach acid.
CaCO3(s)+ 2HCl(aq)= CO2(g)+ CaCl2(aq)+ H20(l)

  • Chemistry -

    "Calculate the number of WHAT of CaCO3 needed....."
    mols HCl = M x L = ?
    mols CaCO3 = 1/2 that (from the coefficients in the balanced equation).
    If you want g CaCO3, then g = mols CaCO3 x molar mass CaCO3.

  • Chemistry -

    multiply ch22+kks2=________________ (G)
    THANK YOU!

  • Chemistry -

    The answer is 1.3 grams of CaCO3. I am not getting that by multiplying mols CaCO3 x molar mass CaCO3.

  • Chemistry -

    Then you must be pushing the wrong buttons on the calculator.
    mols HCl = M x L = 0.1 x 0.250 = 0.0125.
    g CaCO3 = 0.0125 x 100.09 = 1.251 g which I would round to 1.2 g; apparently the author of the problem rounds another way to obtain 1.3 g.
    I don't understand, if you tried to work this yourself, how you ended up with anything different.

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