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Question

A 0.972 g sample of CaCl₂-2H₂O and K₂C₂O₄-H₂O solid salt mixture is dissolved in 150 mL of deionized water, previously adjusted to a pH that is basic. The precipitate, after having been filtered and air-dried, has a mass of 0.375 g. The limiting reactant in the salt mixture was later determined to be CaCl₂-2H₂O.

q1- How many grams of the excess reactant, K₂C₂O₄-H₂O reacted in the mixture?

q2-What is the % by mass of CaCl₂-2H₂O?

q3-How many germs of the K₂C₂O₄-H₂O in the sat mixture remain unaffected?

 

 

Answer 1

CaCl2(aq) + K2C2O4(aq) ===> 2KCl(aq) + CaC2O4(s)  note: calcium oxalate is insoluble, this the rx takes place.

mass CaC2O4 produced = 0.375 g

moles CaC2O4 produced = 0.375 g x 1 mole/128 g = 0.00293 moles

moles K2C2O4-H2O reacted = 0.00293 (note 1:1 mole ratio K2C2O4:CaC2O4)

grams K2C2O4-H2O reacted = 0.00293 moles K2C2O4-H2O x 184 g/mole = 0.539 g reacted 

 

moles CaCl2-2H2O initially present = 0.00293 moles (note 1:1 mole ratio CaCl2:CaC2O4)

mass CaCl2-2H2O initially present = 0.00293 moles x 147 g/mole = 0.431 g

% by mass CaCl2-2H2O = 0.431 g/0.972 g (x100%) = 44.3%

 

grams K2C2O4-H2O that remain unaffected

mass of sample = 0.972 g 

mass of CaCl2-2H2O = 0.431 g

mass of K2C2O4-H2O = 0.972 g - 0.431 g = 0.541 g 

grams K2C2O4-H2O remaining unaffected = 0.541 g - 0.539 g = 0.002 g