Given these chemical reactions:

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).

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.

fjdksal;

To calculate the masses of Ca(NO3)2·4H2O(s) and KIO3(s) required to make 10 g of Ca(IO3)2(s), we need to follow these steps:

Step 1: Calculate the molar mass of Ca(IO3)2:
The molar mass of Ca(IO3)2 can be calculated by adding the molar masses of calcium and iodate ions.
Ca: 40.08 g/mol
IO3: 126.90 g/mol

Molar mass of Ca(IO3)2 = (40.08 g/mol) + 2(126.90 g/mol) = 293.88 g/mol

Step 2: Calculate the number of moles of Ca(IO3)2:
Using the molar mass calculated in step 1, we can calculate the number of moles of Ca(IO3)2 in 10 g of the compound.
Number of moles = Mass / Molar mass
Number of moles = 10 g / 293.88 g/mol = 0.034 moles

Step 3: Determine the limiting reagent:
To determine the limiting reagent, we need to compare the stoichiometry of the chemical reactions given. From the first reaction, we can see that 1 mole of Ca(IO3)2 is formed from 1 mole of Ca^(2+) and 2 moles of IO3^(-). Therefore, we need 1 mole of Ca^(2+) and 2 moles of IO3^(-) to produce 1 mole of Ca(IO3)2.

From the second reaction, we can see that 1 mole of Ca^(2+) is formed from 1 mole of Ca(NO3)2·4H2O. Therefore, to determine the amount of Ca(NO3)2·4H2O needed, we need to compare the moles of Ca^(2+) required with the amount of Ca^(2+) we have.

Step 4: Calculate the moles of Ca(NO3)2·4H2O needed:
Since the Ca^(2+) concentration needs to be in excess by 20%, we need to calculate the minimum moles needed and then increase it by 20%.

From the first reaction, we know that 1 mole of Ca(IO3)2 requires 1 mole of Ca^(2+). Therefore, the minimum moles of Ca^(2+) needed are 0.034 moles.
The 20% excess is calculated by multiplying the minimum moles by 1.2 (or adding 20% to it).
Moles of Ca^(2+) needed = 0.034 moles * 1.2 = 0.0408 moles

From the second reaction, we know that 1 mole of Ca(NO3)2·4H2O produces 1 mole of Ca^(2+). So, the moles of Ca(NO3)2·4H2O needed will be the same as the moles of Ca^(2+) needed, which is 0.0408 moles.

Step 5: Calculate the mass of Ca(NO3)2·4H2O:
To calculate the mass of Ca(NO3)2·4H2O needed, we need to multiply the moles of Ca(NO3)2·4H2O by its molar mass.

Molar mass of Ca(NO3)2·4H2O = (40.08 g/mol + 2(14.01 g/mol + 3(16.00 g/mol))) + 4(1.01 g/mol + 2(16.00 g/mol)) = 236.15 g/mol

Mass of Ca(NO3)2·4H2O = Moles * Molar mass
Mass = 0.0408 mol * 236.15 g/mol = 9.6336 g

Therefore, to produce 10 g of Ca(IO3)2, you would need approximately 9.63 grams of Ca(NO3)2·4H2O.