Calculate the masses of Ca(NO3)2•4H2O(s) and KIO3(s) required to make 10.0 g of

Question

Calculate the masses of Ca(NO3)2•4H2O(s) and KIO3(s) required to make 10.0 g of Ca(IO3)2(s).

Calculation for the required mass of KIO3(s)

Calculation for the required mass of Ca(NO3)2•4H2O(s)

Describe how you would make the three calcium solutions (solutions A, B, and C).Include glassware.

HELP!!! ---these below will help the calcualtions

Calculate the masses of Ca(NO3)2•4H2O(s) and KIO3(s) required to make 10.0 g of Ca(IO3)2(s). Calculate the amount of Ca(NO3)2 •4H2O(s) needed so that the Ca2+ 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 10.0 g of calcium iodate). Thoroughly document all of your calculations for both Ca2+ and IO3

Calculate the mass of Ca(NO3)2 •4H2O(s) required to generate a 0.1XX M Ca 2+ solution in a 100 mL volumetric flask

It is written as 0.1XX M because you don’t know the actual concentration of

calcium until you weigh the solid and generate the solution, but here is what you do know: (a.) it

should be close to 0.1 M; (b.) the final concentration should be known out to 3 significant figures;

and (c.) you should use the actually generated concentration for all of your subsequent

calculations (not just 0.1 M).

Explanation / Answer

Ca+2 + 2IO3- --------------------> Ca(IO3)2

moles Ca (IO3)2 = 10 / 389.88 = 0.02565

moles of Ca(IO3)2 = Ca (NO3)2 .4H2O

moles = mass x molar mass

mass of Ca (NO3)2 .4H2O = 0.02565 x 236.15 = 6.057 g

with 20% excess means = 1 + (20 / 100 ) = 1.2

with this excess = 6.057 x 1.2 = 7.27 g Ca (NO3)2 .4H2O

Ca (NO3)2 .4H2O mass = 7.27 g

moles of KIO3 = 2 x 0.02565 = 0.0513

mass = 0.513 x 214 = 10.98g

mass of KOI3 = 10.98 g

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