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?

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.

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

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

To answer these questions, we need to understand the concepts of stoichiometry and limiting reagents.

(a) The molecular equation for the reaction can be determined based on the reactants involved. From the information provided, we know that the reactants are CaCl2 · 2H2O and K2C2O4 · H2O. The balanced molecular equation would be:

CaCl2 · 2H2O + K2C2O4 · H2O → CaC2O4 + KCl + 3H2O

(b) The net ionic equation shows only the species that actively participate in the reaction. To write the net ionic equation, we need to identify the compounds that dissociate into ions in the reaction. In this case, both CaCl2 · 2H2O and K2C2O4 · H2O dissociate into ions. Therefore, the net ionic equation is:

Ca2+ + 2Cl- + 2H2O + 2K+ + C2O4^2- → CaC2O4 + 2K+ + 2Cl- + 3H2O

(c) The moles of CaCl2 · 2H2O can be calculated using the mass of the precipitate and its molar mass.

Molar mass of CaCl2 · 2H2O = 111.0 g/mol
Mass of precipitate = 0.284 g

Moles = Mass / Molar mass
Moles of CaCl2 · 2H2O = 0.284 g / 111.0 g/mol

(d) The grams of CaCl2 · 2H2O reacted can be calculated using the moles of CaCl2 · 2H2O and its molar mass.

Grams = Moles * Molar mass

(e) Similarly, the moles of K2C2O4 · H2O can be calculated using the mass of the precipitate and its molar mass.

Molar mass of K2C2O4 · H2O = 245.1 g/mol
Mass of precipitate = 0.284 g

Moles = Mass / Molar mass

(f) The grams of K2C2O4 · H2O reacted can be calculated using the moles of K2C2O4 · H2O and its molar mass.

Grams = Moles * Molar mass

(g) To find the grams of K2C2O4 · H2O remaining unreacted, we need to subtract the grams of K2C2O4 · H2O reacted from the initial amount of K2C2O4 · H2O in the salt mixture. Since the information provided does not specify the initial amounts, this calculation cannot be performed based on the given data.

(h) To determine the percent by mass of each salt in the mixture, we need to know the masses of each salt in the initial mixture. As mentioned above, the provided information does not include these values, so calculating the percent by mass is not possible based on the given data.

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.