A. Precipitation of CaC,O H,O from the Salt Mixture Unknown number 1. Mass of be

Question

A. Precipitation of CaC,O H,O from the Salt Mixture Unknown number 1. Mass of beaker (g) 2. Mass of beaker and salt mixture (g) 3. Mass of salt mixture (g) 4. Mass of filter paper (8) 5. Mass of filter paper and product after air-dried Trial 1 .oos 3.S770 2.9701 or oven-dried () 6. Mass of dried product (g) 7. Formula of dried product B. Determination of Limiting Reactant 1. Limiting reactant in salt mixture (write complete formula) 2. Excess reactant in salt mixture (write complete formula) Data Analysis 1 Moles of Cac,o,H,O (or Cac oJ precipitated (mob) 2. Moles of limiting reactant in salt mixture (mol) .formula of limiting hydrate C) 4. Mass of excess reactant in salt mixture (8) Mass of limiting reactant in salt mixture (g) formula of excess hydrate 5. Percent limiting reactant in salt mixture (%) 6. Percent excess reactant in salt mixture (%) 7. Mass of excess reactant that reacted (g) 8. Mass of excess reactant, unreacted (a) -Show calculations for Trial 1 on next page.

Explanation / Answer

The reaction involed in the experiment is CaCl2(aq) + K2C2O4•H2O (aq) ? CaC2O4•H2O (s) + 2 KCl (aq)

So per the reaction, CaC2O4•H2O would precipitate and KCl would remain in the folterate. So

7. Formula of the dried product is CaC2O4•H2O

Data Analaysis:

1. Moles of CaC2O4•H2O precipitated = gms of CaC2O4•H2O / molar mass of CaC2O4•H2O

Moles of CaC2O4•H2O precipitated = 2.9704 g / 146.1123 g/mol = 0.02330 moles

2. Per the reaction, for every 1 mole of mole of CaC2O4•H2O 1 mole of K2C2O4•H2O would be produced.

Formula of limiting reactant: K2C2O4•H2O

So, Moles of limiting reactant K2C2O4•H2O in the salt mixture = moles of CaC2O4•H2O = 0.02330 moles

3. Mass of K2C2O4•H2O in the salt mixture = moles * Molar mass of K2C2O4•H2O = 0.02330 * 184.2309 = 3.7453 g

4. Mass of excess reactant in the salt mixture = moles * Molar mass of CaCl2

Per the reaction, Moles of excess reactant CaCl2 = moles of CaC2O4•H2O = 0.02330 moles

Mass of CaCl2 in the salt mixture = moles * Molar mass of CaCl2 = 0.02330 moles * 110.98 g/mol = 2.2562 g

5. %limiting reactant in salt mixture = mass K2C2O4•H2O/mass salt mix * 100% = 3.7453g/1.005g * 100% = 372.2 %

The % is above 100 as the product woudn't have been dried completely.

6. %Excess reactant in salt mixture = mass of CaCl2/mass of salt mix * 100% = 2.2562 g/1.005g * 100% = 324.5 %

The % is above 100 as the product woudn't have been dried completely.

7. Mass of Excess reactant, reacted = This value cannot be calculated because we have a product thats not dried completely.

8. Mass of Excess reactant, unreacted = This value cannot be calculated because we have a product thats not dried completely.

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