A 0.879g sample of a CaCl 2 . 2H 2 O/K 2 C 2 O 4 .H 2 O solid salt mixture is di

Question

A 0.879g sample of a CaCl2. 2H2O/K2C2 O4.H2O solid salt mixture is dissolved in approximately 150mL of deionized water, previously adjusted to a PH that is basic. The precipitate, after having been filtered and air dried, has a mass of 0.284 g. The limiting reactant in the salt was later determined to be CaCl2. 2H2O.

Write a molecular form of the equation for the reaction

Write the net ionic equation for the reaction.

How many moles and grams of CaCl2. 2H2O reacted in the reaction mixture.?

How many moles and grams of the excess reactant, K2C2 O4.H2O, reacted in the mixture?

How many grams of K2C2 O4.H2O in the salt mixture remain unreacted (in excess)?

What is the percent by mass of each salt in the mixture?

Explanation / Answer

A 0.879g sample of a CaCl2. 2H2O/K2C2 O4.H2O solid salt mixture is dissolved in approximately 150mL of deionized water, previously adjusted to a PH that is basic. The precipitate, after having been filtered and air dried, has a mass of 0.284 g. The limiting reactant in the salt was later determined to be CaCl2. 2H2O.

1. Write a molecular form of the equation for the reaction

CaCl2.2H2O(aq) + K2C2O4.H2O(aq) ----> CaC2O4(s) + 2 KCl(aq) + 3 H2O(l)

2. Write the net ionic equation for the reaction.

Ca2+(aq) + (C2O4)2-(aq) ---> CaC2O4(s)

3. How many moles and grams of CaCl2. 2H2O reacted in the reaction mixture.?

The moles of CaC2O4 formed will be = mass of precipitate / molecular weight of CaC2O4

= 0.284 / 128 = 0.0022 moles

As given that the calcium chloride dihydrate is the limiting reagent , so the moles of calcium chloride dihdyride will be 0.0022 moles

So mass of calcium dihydride = moles X molecular weight = 0.0022 X 147g / mole = 0.323 grams

4. How many moles and grams of the excess reactant, K2C2 O4.H2O, reacted in the mixture

So mass of potassium oxalate will be = 0.879 - 0.323 = 0.556 grams

So moles of potassium oxalaate will be = mass / molecualr weight = 0.556 / 184 = 0.0030

5. Moles of K2C2 O4.H2O reacted will be 0.0022

So moles left = 0.0030 - 0.0022 = 0.0008 moles

so mass of K2C2 O4.H2O left = moles X molecular weight = 0.0008 X 184 = 0.1472 grams

6. Percentage mass will be

MAss of K2C2 O4.H2O / total mass = 0.556 X 100 / 0.879 = 63.25 %

MAss of CaCl2. 2H2O / total mass = 0.323 X 100 / 0.879 = 36.75 %

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