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17 April, 13:33

A 1.248 g sample of limestone rock is pulverized and then treated with 30.00 ml of 1.035 m hcl solution. the excess acid then requires 11.56 ml of 1.010 m naoh for neutralization. part a calculate the percent by mass of calcium carbonate in the rock, assuming that it is the only substance reacting with the hcl solution.

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  1. 17 April, 13:37
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    First, we will find the moles of HCl entered. This will require the use of the formula:

    Moles = molarity * volume (in L)

    Moles = 1.035 * 0.03

    Moles of HCl = 0.0315

    Next, we find the amount reacted with NaOH. Note that each mole of NaOH reacts with one mole of HCl (determined from reaction equation)

    Moles of NaOH = 1.01 * 0.01156

    Moles of NaOH = 0.01168

    Moles of HCl reacted with calcium carbonate = 0.0315 - 0.01168

    = 0.01982 mole

    The reaction equation is:

    CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O

    So the moles of CaCO₃ were half that of HCl, which means:

    0.01982 / 2 = 0.00991

    The mass of calcium carbonate is:

    Mass = moles * molecular weight

    Mass = 0.00991 * 100

    Mass = 0.991 grams

    Percentage mass = (0.991 / 1.248) * 100

    The ore has 79.41% calcium carbonate by mass.
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