DUE TOMORROW PLEASE HELP! A sample of a CaCl2 middot 2H2O/K2C2O4 middot H2O soli
ID: 850133 • Letter: D
Question
DUE TOMORROW PLEASE HELP!
A sample of a CaCl2 middot 2H2O/K2C2O4 middot H2O solid salt mixture is dissolved in ap 150 mL of deionized water previously adjusted to a pH that is basic. The precipitate, after having been filtered, was air-dried and weighed. Data for Trial 1 were obtained as shown. Complete the following table. (See Report Sheet.) Record calculated values with the correct number of significant figures. Precipitation of CaC2O4 middot H2O from the Calculation Zone Salt Mixture Mass of salt mixture (g) 0.879 Data Analysis, 1. Mass of filter paper (g) 1.896 Mass of filter paper and CaC2O4 middot H2O (g) 2.180 Mass of air-dried CaC2O4 middot H2O (g) Determination of Limiting Reactant Limiting reactant in salt mixture CaCl2 middot 2H2O Data Analysis, 3. Excess reactant in salt mixture Data Analysis Moles of CaC2O4 middot H2O precipitated (mol) Show calculation. Data Analysis, 4. Moles of limiting reactant in salt mixture (mol) See equation 8.1. Mass of limiting reactant in salt mixture (g) Show calculation. Mass of excess reactant in salt mixture (g) Equals mass of salt mixture minus mass of limiting reactant. Show calculation. Data Analysis, 5. Percent limiting reactant in salt mixture (%) Show calculation. Percent excess reactant in salt mixture (%)Explanation / Answer
A. Mass of CaC2O4.H2O = 2.180 - 1.896 = 0.284g
B. excess reactant = K2C2O4.H2O
Data analysis:
1. moles of precipitate = 0.284/ Mol. weight = 0.284/146 = 0.00194 moles
3. Assuming initiallly both salts were mixed in 1:1 ratio.
Total original mass of salt mixture= 0.879g
Thus mass of each salt = 0.879/2 = 0.4395 g
Remaining mass of limiting reagent = 0.4395 - 0.284 = 0.1555 g
2. no. of moles of limiting reagent = 0.1555/146 = 0.001065 moles
4. mass of excess reagent = 0.4395g Since we are assuming K2C2O4.H2O doesn't precipitate.
5. % limiting reagent = 0.1555/ (0.4395+0.1555) = 0.261 = 26.1%
6. % excess = 100 -26.1 = 73.9%
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