Answer:
a. CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)
b. Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)
C. moles of CaCl₂.2 H₂O reacted in the mixture = 0.00222 moles
d. Mass of CaCl₂.2 H₂O reacted = 0.326 g
e. Moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles
f. Mass of K₂C₂O₄.H₂O reacted = 0.408 g
g. mass of K₂C₂O₄.H₂O remaining unreacted = 0.145 g
h. Percent by mass CaCl₂.2 H₂O = 37.1%
Percent by mass of K₂C₂O₄.H₂O = 62.9%
Explanation:
a. Molecular equation of the reaction is given below :
CaCl₂.2 H₂O (aq) + K₂C₂O₄. H₂O (aq) ----> 2 KCl + CaC₂O₄ (s) + 3 H₂0 (l)
b. The net ionic equation is given below
Ca²+ (aq) + C₂O₄²- (aq) ----> CaC₂O₄ (s)
C. mass CaC₂O₄ produced = 0.284 g, molar mass of CaC₂O₄ = 128 g/mol
moles CaC₂O₄ produced = 0.284 g / 128 g/mol = 0.00222 moles
Mole ratio of CaC₂O₄ and CaCl₂.2 H₂O is 1 : 1, therefore moles of CaCl₂.2 H₂O 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 CaCl₂.2 H₂O = 147 g/mol
Mass of CaCl₂.2 H₂O reacted = 0.00222 moles × 147 g/mol = 0.326 g
e. Mole ratio of K₂C₂O₄.2 H₂O and CaC₂O₄ is 1 : 1, therefore, moles of K₂C₂O₄.2 H₂O reacted = 0.00222 moles
f. Mass of K₂C₂O₄.H₂O reacted in the mixture = number of moles × molar mass
Molar mass of K₂C₂O₄.H₂O = 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 CaCl₂.2 H₂O in sample completely used up = 0.326 g
mass of K₂C₂O₄.H₂O in sample = 0.879 g - 0.326 g = 0.553 g
mass of K₂C₂O₄.H₂O remaining unreacted = 0.553 g - 0.408 g = 0.145 g
h. Percent by mass CaCl₂.2 H₂O = 0.326 /0.879 x 100% = 37.1%
Percent by mass of K₂C₂O₄.H₂O = 0.553/0.879 × 100% = 62.9%
