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

    What mass of CaCO3 would you need to completely neutralize 50ml of 0.1 M HCl?

    Suppose a chemist mixed 2.00g of CaCO3 (s) with 100ml of 0.250 M HCl. What mass of each of the three products would form?

    In the question above, what is the concentration of CaCl2 in solution after that reaction has been completed?

    Please put in terms I can understand.

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    Solution Preview

    First of all, balance the reaction before you start to solve anything (put a 2 in front of the HCl)...

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

    PART 1

    50 mL is 0.050 L

    moles of HCl = (0.050 L)(0.1 M) = 0.005 mol HCl

    For every 2 mol of HCl you need one mole of CaCO3, therefore

    moles of CaCO3 = (1 mol CaCO3/ ...

    Solution Summary

    The concentration and mass of a chemist function is provided. The solution is completed step-by-step.