A. Precipitation of CaC,O H,O from the Salt Mixture 3. Mass of salt misture ) 4.

Question

A. Precipitation of CaC,O H,O from the Salt Mixture 3. Mass of salt misture ) 4. Mass of fiher paper () 5. Mass of filter paper and CaC,0.110O 8) 6. Mass of air dried CaC O H,O (8) Data Analysis, 1. 7.130. 130- 1,896-,334 B. Determination of Limiting Reactant Data Analysis, 3. CaCl-2H O I. Limiting reactant in salt mixture 2. -Excess reactant in salt mixture Data Analysis 1: Moles of CaC,O11,0 precipitated (mol) Data Analysis, 4 Show calculation 2. Moles of limiting reactant in salt mixture (mol) See equation 8.1 Show calculation Equals mass of silt misture minus mass of limiting reictant. 3. Mass of limiting reactant in salt mixture (g) 4. Mass of excess reactant in salt mixture (8) Data Analysis, 5 Show calculation. 5. Percent limiting reactant in salt mixture (%) Show calculation 6. Percent excess reactant in salt mixture (%)

Explanation / Answer

CaCl22H2O(aq) + K2C2O4H2O(aq) = CaC2O4H2O(s) + 2KCl(aq) + 2H2O(l)

1. Molar Mass of CaC2O4H2O(s) = 146.1123 g/mol

so, 0.284 gm = 0.00194 mole

2. moles of limiting reactant = 0.00194 mole

3.CaCl22H2O = 147.02 gm/mole

so, 0.00194 mole = 0.2852 gm

4.

0.879 gm - 0.2852 gm = 0.5938 gm

5.

(0.2852/ 0.879)*100 = 32.44%

6.

(0.5938/ 0.879)*100 = 67.56%

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