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A 0.879 g sample of a CaCl2 ∙ 2 H2O / K2C2O4 ∙ H2O solid salt mixture is dissolved in 150 mL of deionized water. A precipitate forms which is then filtered and dried. The mass of this precipitate is 0.284 g. The limiting reagent in the salt mixture was later determined to be CaCl2 ∙ 2 H2O.

a. Write the molecular form of the equation for the reaction.
b. Write the net ionic equation for the reaction
c. How many moles of CaCl2 ∙ 2 H2O reacted in the reaction mixture?
d. How many grams of CaCl2 ∙ 2 H2O reacted in the reaction mixture?
e. How many moles of K2C2O4 ∙ H2O reacted in the reaction mixture?
f. How many grams of K2C2O4 ∙ H2O reacted in the reaction mixture?
g. How many grams of K2C2O4 ∙ H2O in the salt mixture remain unreacted?
h. What is the percent by mass of each salt in the mixture?

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Answer:

a. CaClâ‚‚.2 Hâ‚‚O (aq) + Kâ‚‚Câ‚‚Oâ‚„. Hâ‚‚O (aq) ----> 2 KCl + CaCâ‚‚Oâ‚„ (s) + 3 Hâ‚‚0 (l)

b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. moles of CaClâ‚‚.2 Hâ‚‚O reacted in the mixture = 0.00222 moles

d. Mass of CaClâ‚‚.2 Hâ‚‚O reacted = 0.326 g

e. Moles of Kâ‚‚Câ‚‚Oâ‚„.2 Hâ‚‚O reacted = 0.00222 moles

f. Mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O reacted = 0.408 g

g. mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O remaining unreacted = 0.145 g

h. Percent by mass CaClâ‚‚.2 Hâ‚‚O = 37.1%

Percent by mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O = 62.9%

Explanation:

a. Molecular equation of the reaction is given below :

CaClâ‚‚.2 Hâ‚‚O (aq) + Kâ‚‚Câ‚‚Oâ‚„. Hâ‚‚O (aq) ----> 2 KCl + CaCâ‚‚Oâ‚„ (s) + 3 Hâ‚‚0 (l)

b. The net ionic equation is given below

Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)

C. mass CaCâ‚‚Oâ‚„ produced = 0.284 g, molar mass of CaCâ‚‚Oâ‚„ = 128 g/mol

moles CaCâ‚‚Oâ‚„ produced = 0.284 g / 128 g/mol = 0.00222 moles

Mole ratio of CaCâ‚‚Oâ‚„ and CaClâ‚‚.2 Hâ‚‚O is 1 : 1, therefore moles of CaClâ‚‚.2 Hâ‚‚O reacted in the mixture = 0.00222 moles

d. Mass of CaCl₂.2 H₂O reacted in the mixture = number of moles × molar mass

Molar mass of CaClâ‚‚.2 Hâ‚‚O = 147 g/mol

Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g

e. Mole ratio of Kâ‚‚Câ‚‚Oâ‚„.2 Hâ‚‚O and CaCâ‚‚Oâ‚„ is 1 : 1, therefore, moles of Kâ‚‚Câ‚‚Oâ‚„.2 Hâ‚‚O reacted = 0.00222 moles

f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass

Molar mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O = 184 g/mol

grams K2C2O4-H2O reacted = 0.00222 moles 184 g/mole = 0.408 g

g. Mass of sample = 0.879 g

mass of CaClâ‚‚.2 Hâ‚‚O in sample completely used up = 0.326 g

mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O in sample = 0.879 g - 0.326 g = 0.553 g

mass of Kâ‚‚Câ‚‚Oâ‚„.Hâ‚‚O remaining unreacted = 0.553 g - 0.408 g = 0.145 g

h. Percent by mass CaClâ‚‚.2 Hâ‚‚O = 0.326 /0.879 x 100% = 37.1%

Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%