te 09 1. Mass of beaker (g) 2. Mass of beaker and salt mixture (g) 3. Mass of sa

Question

te 09 1. Mass of beaker (g) 2. Mass of beaker and salt mixture (g) 3. Mass of salt mixture (g) 4. Mass of filter paper (g) 5. Mass of filter paper and product after air-dried 2.13 or oven-dried (g) 6. Mass of dried product (g) 7. Formula of dried product 221 Determination of Limiting Reactant 1. Limiting reactant in salt mixture (write complete formula) 2. Excess reactant in salt mixture (write complete formula) B. 1c29 Data Analysis 1. Moles of CaC20,-H20 (or CaC04) precipitated (mol) 2. Moles of limiting reactant in salt mixture (mol) of limiting hydrate o-7fformula 3. Mass of limiting reactant in salt mixture (g) 4. Mass of excess reactant in salt mixture (g) .formula of excess hydrate 5. Percent limiting reactant in saltmixture(%) 6-percent excess reactant insalt mixture (%) 7. Mass of excess reactant that reacted () 8. Mass of excess reactant, unreacted (g) EXP -(eat Show calculations for Trial 1 on next page.

Explanation / Answer

Trial 1

Mass of beaker (g) = 47.770g

Mass of beaker + salt mixture = 48.241 g

Mass of salt mixture (g) = 1.001 g

Mass of filter paper (g) = 1.907 g

Mass of filter paper and product = 2.134 g

Mass of dried product = 0.227 g

Data analysis

The reaction is CaCl2 (aq) + K2C2O4 (aq) ----> CaC2O4(s) + 2KCl(aq)

1) Mole of moles of CaC2O4 precipitated = weight of CaC2O4/Molar mass of CaC2O4

0.227 g/128.097 g/mol = 0.00177 moles

2) Mole of is limiting reactant in salt mixture = 0.00177 moles

Formula of limiting reactant is K2C2O4.H2O

3) Mass of is limiting reactant in salt mixture = moles x molar mass

Molar mass of K2C2O4 is 184.2309g/mol

Mass of is limiting reactant in salt mixture = 0.00177 moles x 184.2309g /mol = 0.326 g

4) Mass of excess reagent in salt mixture = Mass of salt mixture - Mass of is limiting reactant

Mass of excess reagent in salt mixture = 1.001 g - 0.326 g = 0.675 g

Formula of excess reagent CaCl2.2H2O

5) Percent of limiting reactant in salt mixture = (0.326 g/1.001g) x 100 = 32.57%

6) Percent of excess reactant in salt mixture = (0.675 g/1.001g) x 100 = 67.43%

7) Mass of excess reactant that reacts

1 mol of K2C2O4.H2O reacts with 1 mol of is CaCl2 (aq)

0.00177 Moles of K2C2O4.H2O reacts with 0.00177 moles of is CaCl2 (aq)

So, moles of excess reactant that reacts = 0.00177 moles

Molar mass of excess hydrate =147.0146 g/mol

Now multiply moles of excess reactant that reacts by molar mass of excess hydrate

Mass of excess reactant that reacts = 0.00177moles x 147.0146 g/moles = 0.260 g

8) Mass of excess reactant unreacted = Mass of excess reagent - Mass of excess reactant that reacts

Mass of excess reactant unreacted = 0.675 g - 0.260 g = 0.415 g

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