Given these chemical reactions:

Question

Ca^(2+)(aq) + 2IO3^(-)(aq) → Ca(IO3)2(s)

Ca(NO3)2·4H2O(s) → Ca^(2+)(aq) + 2NO3^(-)(aq) + 4H2O(l)

KIO3(s) → K^(+)(aq) + IO3(-)(aq)

Calculate the masses of Ca(NO3)2·4H2O(s) and KIO3(s) required to make 10 g of Ca(IO3)2(s). This is done by combining IO3^(-) with a slight excess of Ca^(2+) and allowing the first reaction above to proceed. Calculate the amount of Ca(NO3)2·4H2O(s) needed so that the Ca^(2+) is in excess of the iodate ion concentration by 20% (i.e., calculate a weight of calcium nitrate that is 20% higher than the minimum required to produce 10 g of calcium iodate).

Answer 1

I would add the two equations needed together to achieve the final one desired. I think doing it piecemeal just complicates the process.
Ca(NO3)2.4H2O + 2KIO3 ==> Ca(IO3)2 + 2KNO3 + 4H2O.

1. You want 10 g Ca(IO3)2. Convert that to moles. #moles = grams/molar mass.
2. Using the coefficients in the balanced equation above, convert moles Ca(IO3)2 to moles KIO3.
3.Now convert moles KIO3 to grams KIO3. grams KIO3 = moles KIO3 x molar mass KIO3. That get the minimum amount of KIO3 needed.

4. For the Ca(NO3)2.4H2O part, repeat step 2 to determine moles Ca(NO3)2.4H2O needed. Then repeat step 3 to convert to grams Ca(NO3)2.4H2O. Now multiply by 0.20 and add to the original calculated amount of Ca(NO3)2.4H2O to get the 20% excess.
Post your work if you have problems.

Answer 2

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