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    I'm working on this question: What volume of dilute hydrochloric acid that has the concentration 0.050 mol dm-3 is needed to react with a tablet containing 400 mg of calcium carbonate?

    The first thing I did is wrote a balanced equation as follows (http://ca.answers.yahoo.com/question...0202727AAbwjZd):

    CaCO3(s) + 2HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)

    After that, I calculated the number of moles CaCO3 as follows:
    #moles(CaCO3) = 0.4 g / 100 g mol-1 = 0.004 mol

    Now, based on the stoichiometry of the reaction, the number of moles of HCl will be 2 X 0.004 mol = 0.008 mol

    And, thus, based on this equation: concentration(mol dm-3) = number of moles (mol) / volume (dm3) we can caclulate the volume needed for HCl as follows:
    0.050 mol dm-3 = 0.008 mol / volume (dm3)
    volume = 0.008 mol / 0.050 mol dm-3 = 0.16 dm3

    Is this correct?

    Thanks.
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    (Original post by SWEngineer)
    I'm working on this question: What volume of dilute hydrochloric acid that has the concentration 0.050 mol dm-3 is needed to react with a tablet containing 400 mg of calcium carbonate?

    The first thing I did is wrote a balanced equation as follows (http://ca.answers.yahoo.com/question...0202727AAbwjZd):

    CaCO3(s) + 2HCl(aq) --> CaCl2(aq) + CO2(g) + H2O(l)

    After that, I calculated the [I]number of moles[I] CaCO3 has as follows:
    #moles(CaCO3) = 0.4 g / 100 g mole-1 = 0.004 mol

    Now, based on the stoichiometry of the reaction, the number of moles of HCl will be 2 X 0.004 mol = 0.008 mol

    And, thus, based on this equation: concentration(mol dm-3) = number of moles (mol) / volume (dm3) we can caclulate the volume needed for HCl as follows:
    0.050 mol dm-3 = 0.008 mol / volume (dm3)
    volume = 0.008 mol / 0.050 mol dm-3 = 0.16 dm3

    Is this correct?

    Thanks.
    yeah, i concur
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    Thanks @shengoc. Who else agrees with this solution? :-)
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    Yeah I agree
 
 
 
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