randlemccray4305
13.07.2021 •
Chemistry
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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Ответ:
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%
Ответ: