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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. (2)
(b) Write the net ionic equation for the reaction (3)
(c) How many moles of CaCl2 ∙ 2 H2O reacted in the reaction mixture? (2)
(d) How many grams of CaCl2 ∙ 2 H2O reacted in the reaction mixture? (2)
(e) How many moles of K2C2O4 ∙ H2O reacted in the reaction mixture? (1)
(f) How many grams of K2C2O4 ∙ H2O reacted in the reaction mixture? (1)
(g) How many grams of K2C2O4 ∙ H2O in the salt mixture remain unreacted? (2)
(h) What is the percent by mass of each salt in the mixture?

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4 answers

  1. Oh Crud!!. I just misread C2O4 as CrO4. But that doesn't change any of the numbers or equation. Thanks for pointing that out. My eyesight is getting really bad. Even with proofing I still made the mistake.

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  2. DrBob, The chemistry is correct but, where did the chromate come from? I think you meant potassium oxalate monohydrate. :-)

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  3. Ya got plenty of wins! Not to worry. :-)

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  4. This looks straight forward to me but I'll get you started.
    a. CaCl2.2H2O + K2CrO4 ==> CaCrO4 + 2KCl + 2H2O
    b. Ca^2+ + CrO4^2- ==> CaCrO4
    c. mols CaCrO4 = grams CaCrO4/molar mass CaCrO4 = ?. g = 0.284
    Since 1 mol CaCl2.2H2O = 1 mol K2CrO4 in the reaction, mols CaCl2.2H2O = mols CaCrO4
    d. g CaCl2.2H2O = mols CaCl2.2H2O x molar mass CaCl2.2H2O = ?
    e. You see how it's done. I'll let you finish. Post your work if you get stuck.

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