show calculation of 1. Moles of CaC2O4 *H2O (CaC2O4) Precipitated (mol) 2 Moles

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show calculation of

1. Moles of CaC2O4 *H2O (CaC2O4) Precipitated (mol)

2 Moles of limiting reactant in salt mixture (mol) and the formula of limiting hydrate

3. Mass of excess reactant in salt mixture (g)

4. Mass of excess reactant in salt mixture (g)
and the 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. (g)

ment 8 Report Sheet Limiting Reactant Desk No Date A. Precipitation of CaC,0,H,O from the Salt Mixture Unknown number SC 1. Mass of beaker (g) Trial I Trial 1 182 05.32 106-92G 2. Mass of beaker and salt mixture (g) 3. Mass of salt mixture (g) 1.094 4. Mass of filter paper (g) 5. Mass of filter paper and product after air-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) O.G39 or oven-dried (g) 2. Data Analysis 1. Moles of CaC2O'H20 (or CaC,04) precipitated ('no) 2. Moles of limiting reactant in salt mixture (mol) .formula of limiting hydrate 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 salt mixture (%) 6. Percent excess reactant in salt mixture (%) 7. Mass of excess reactant that reacted (g) 8. Mass of excess reactant, unreacted (g) Show calculations for Trial 1 on next page. Experiment 8 131

Explanation / Answer

1. Moles of CaC2O4 *H2O (CaC2O4) Precipitated (mol)?

From given data, mass of dried product (CaC2O4) is 0.252 g, 0.347 g

Molecular weight of CaC2O4= 146.111g/mol

moles (n) = mass/m.wt

            n = 0.252/146.111 = 1.7247 * 10-3

            n = 0.347/146.111 = 2.3749 * 10-3

        

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