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please check if I did this right ...

calculate the mass of white solid CaCO3 that forms when 98ml of a 0.1M Ca(NO3)2 solution is mixed with 54 ml of a 2.5M Na2CO3 solution.

Ca(NO3)2 + 2NaCO3 -> Ca(CO3)2 + 2NaNO3

.098*0.1 =.0090
.054*2.5 =.135

limiting reactant is Ca(NO3)2

.0090 * 160.1 = 1.44g

  • chemistry -

    the balanced equation should be this:

    Ca(NO3)2+ Na2CO3>>CaCO3 + 2NaNO3

    You had two formulas wrong.

  • chemistry -

    Thank you

    so with those corrections my new calculations would be

    .0090*100.05 = .90045g

    Hopefully this time it is right.

    Obviously I am not chemisrty gifted.

  • chemistry -

    so with those corrections my new calculations would be

    .0090*100.05 = .90045g
    The 0.0090 number you have comes from 0.098 x 0.1 = 0.0098 and you made a typo when you entered it. Then 0.0098 x 100.05 = 0.98 g

  • chemistry -

    i have a question of why did you put 2 after the parentheses Ca(CO3)2

    isn't Ca has the charge number of 2+
    and the CO3 has the charge of negative 2

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