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math Steve helps me

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1.64g of a mixture of cac03 and mgc03 was dissolved in 50ml of 0.8 m Hcl. The excess of acid required 16ml of 0.25< NAOH for neutralization. Calculate the percentage of cac03 and mgc03 in the sample

The right answer mgc03 52.02% and caco 47.98%

You need to use 3 solution of drug H containing respectively 50%, 20% and 5% to make 600ml of 10% drug H. How much of the 50% will you use?

  • math Steve helps me - ,

    I assume that 0.8m actually is 0.8M.
    Typical equation is
    CaCO3 + 2HCl ==> CaCl2 + H2O + CO2
    HCl used initially = M x L = 0.8M x 0.05L = 0.04 mols to begin.
    Back titrated = M x L = 0.25M x 0.016L = 0.004 mols excess HCl.
    Amt HCl used = 0.04mols - 0.004 mols = 0.036 mols. Convert that to mols MgCO3 and CaCO3 = 1/2 x 0.036 = 0.018 mols used for the 1.64 g of the mixture.

    Let X = g CaCO3
    and Y = g MgCO3
    ---------------
    equation 1 is X + Y = 1.64g
    equation 2 is
    (X/molar mass CaCO3) + (Y/molar mass MgCO3) = 0.018
    ------------------------
    Solve those two equations simultaneously to find X and Y.
    Then (X/1.64)*100 = %CaCO3 and
    (Y/1.64)*100 = %MgCO3

    #2
    mL x % = mL x %
    XmL x 50% = 600 mL x 10%.
    Solve for X.
    Note: You don't need any of the 20% and you don't need any of the 5%.

  • math Steve helps me - ,

    From the previous solution, we ended up with

    .776g CaCO3 = .00775 moles
    .018-.00775 = .0102 moles MgCO3

    By mass, that's

    .00775*100.09=.7756g CaCO3
    .0102*84.31 = .8599g MgCO3
    total mass: 1.6355g
    47.4% CaCO3

    That's pretty close, eh? Maybe roundoff errors make up the small difference.

    Looks like you need to study how mixture problems work.

    There are many solutions to the drug H problem, since you don't specify any other conditions other than the total volume and concentration.

    x+y+z = 600
    .05x + .20y + .50z = .10*600

    Lots of solutions to that exist:

    5% 20% 50%
    440 140 20
    500 50 50
    520 20 60

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