PRIOR To cLAss CALcULATION: Calculate the masses of ca (NOs)2.4H20 and 3(s) requ

Question

PRIOR To cLAss CALcULATION: Calculate the masses of ca (NOs)2.4H20 and 3(s) required to make 10.0 g of Ca(o, hay. Calculate the amount of Ca(NO, 2.4H20h needed so that the ca 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 Ca" and loa on the laboratory data sheet. Get the approval of the laboratory instructor or TA prior to actually

Explanation / Answer

Write down the balanced chemical equation:

Ca(NO3)2.4H2O (s) + 2 KIO3 (s) -------> Ca(IO3)2 (s) + 2 KNO3 (s) + 4 H2O (l)

As per the stoichiometric equation,

1 mole Ca(NO3)2.4H2O = 2 mole KIO3 = 1 mole Ca(IO3)2

Find out the molar masses:

Ca(NO3)2.4H2O = 236.1489 g/mol

KIO3 = 214.001 g/mol

Ca(IO3)2 = 389.88334 g/mol

Now calculate the masses of Ca(NO3)2.4H2O and KIO3 required to produce 10.0 g Ca(IO3)2.

Ca(NO3)24H2O: [10.0 g Ca(IO3)2]*[1 mole Ca(IO3)2/389.88334 g Ca(IO3)2]*[1 mole Ca(NO3)2.4H2O/1 mole Ca(IO3)2]*[236.1489 g Ca(NO3)2.4H2O/1 mole Ca(NO3)2.4H2O] = 6.0569 g Ca(NO3)2.4H2O (ans)

KIO3: [10.0 g Ca(IO3)2]*[1 mole Ca(IO3)2/389.88334 g Ca(IO3)2]*[1 mole Ca(NO3)2.4H2O/2 mole KIO3]*[214.001 g KIO3/1 mole KIO3] = 5.4888 g KIO3 (ans).

Ofcourse, we require a smaller amount of KIO3 and thus KIO3 will be the limiting reactant.

In the next part, we are required to find the mass of Ca(NO3)2.4H2O that is 20% higher than the mass of KIO3 required in the above step.

Therefore, the mass of Ca(NO3)2.4H2O required = (5.4888 + 20/100*5.4888) g = 6.58656 g (ans).

Suppose we want to prepare a solution that 0.105 M in Ca2+.

Ca(NO3)2.4H2O dissociates completely in aqueous solution as below:

Ca(NO3)2.4H2O (s) -----> Ca(NO3)2.4H2O (aq) ------> Ca2+ (aq) + 2 NO3- (aq) + 4 H2O (l)

As per the balanced stoichiometric reaction,

1 mole Ca(NO3)2.4H2O = 1 mole Ca2+

Moles of Ca(NO3)2.4H2O required to prepare 100 mL of 0.105 M solution = (100 mL)*(1 L/1000 mL)*(0.105 mol/L) = 0.0105 mole.

Mass of Ca(NO3)2.4H2O required = (0.0105 mole)*(236.1489 g/mol) = 2.4796 g (ans).

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.