Question

How do you calculate questions 1 - 8 for the data analysis section?

A. Precipitation of CaC,O, H,O from the Salt Mixture Unknown number Trial 1 Trial 2 1. Mass of beaker (g) 59.45 59.45 5.oS Mass of beaker and salt mixture (g) 2. 3. Mass of salt mixture (g) I, oo oS 4. Mass of filter paper (g 5. Mass of filter paper and product after air-dried or oven-dried (g) 0, 35 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 (owrite complete formula) Data Analysis 1. Moles of CaC,0,H,o (or CaC,O) precipitated (mol) 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 () 8. Mass of excess reactant, unreacted (g) ing only the ions that show evidence of a chemical reaction ie precipitate) and by removing the spectator ions (i.e., no change he reaction), we have the net ionic equation for the observedr the forma- of ionic form ofío tions or anions that Presen in ony observable tion of a mical reoction ate monohydrate is thermally stable below -90 C bur of KCOHO, molar ma during the reacti that Calcium oxalate monohydrate is thermally anhydrous salt. CaC,04, at temperatures above Therefore, one mole of Ca+ (from one m g/mol) reacts with one mole of C,O (from one mo e ions observed chemical sented in O, molar mass = 147.02 le of CaCl, 2H,O, 146.12 g/mot the calcium oxalate is heated to temperatures greater than 110'rdying, then drous CaC20, (molar mass I 28. 10 g/mol) is the product. one mole of C,0,2 (from one mole of K,C,O, H,o 24 gmol) to produce one mole of CaC,O,H,O (molar mass form a heterogeneous mixture of unknown composition. The mass of the solid mi measured and then added to water-insoluble CaC,O:H,O forms. TheCacor is In Part A of this experiment the solid reactant salts CaCl2-2HO and K,C,0-H precipitate is collected via gravity filtration and dried, and its mass is measured The percent composition of the salt mixture is determined by first testir oxalate monohydrate is determined from two precipitation tests of the fi reaction ant. In Part B, the limiting reactant for the formation of solid the limiting react mixture from Part A: (1) the mixture is tested for an excess of calciu ith en gxplato reagent-observed formation of a precipitate indicates the presence of an excess of calcium ion (and a limited amount of oxalate ion) in the salt mit. ture: 12) the mixture is also tested for an excess of oxalate ion with a calcium reagentobserved formation of a precipitate indicates the presence of an excess of oxalate ion (and a Timited amount of calcium ion) in the salt mixture cium The calculations for the analysis of the salt mixture require some attention. "How do l proceed to determine the percent com K-C20,-H20 by measuring only the mass of the CaC2O'H2O precipitate?. position of a salt mixture of CaCl, 2H,0 and Example: A 0.538-g sample of the salt mixture is added to water and after drying (to less than 90°C) 0.194 g of CaC.O4H.O is measured. Tests reveal that KC,O,HO is the limiting reactant. What is the percent composition of the salt mixture? Since K,C,OHO is the limiting reactant, how many grams of the excess CaCl2 2H2O were in the salt mixture? Solution: Since K,C,O,H,0 is the limiting reactant, then, according to equation 8.1, the moles of K,C,O,H,0 in the salt mixture equals the moles of CaC,0:H0 formed. Therefore, the calculated mass of K.C,O, H 0 in the original salt mixture is I mol CaC,O, H O 146.12 g CaC,o., HO 1 mol K,C,O.HO 184.24 K,C,O H,0 l mol CaGO,H20 × l mol KC204-HO) -0.245 g KC,O H,O in the salt mixture. The percent by mass of K,C,O,H,0 in the original salt mixture is %K,C,O/H() = 0.538 g sample -× 100=4 The mass of the CaCl,2H,0 in the salt mixture is the difference between the n of the sample and the mass of K C,o H,0 or (0.538 g 0.245 g-) 0.293 percent by mass of CaCl:2H,0 in the original salt mixture is 0.538 g-0.245 g 2H,00.538 g sample % CaCl- 100-54.5% CaCl2-2H2O Reactant

Answer 1

Data analysis:

1. Moles of CaC₂O₄. H₂O = mass /molar mass = (0.35/146.12)

= 0.00239

2. As limiting reactant is CaCl₂, H₂O

Then, moles of CaC₂O₄ , H₂O = moles of CaCl₂, H₂O = 0.00239

Mass of limiting reactant (CaCl₂, H₂O) = moles of limiting reactant *molar mass of limiting reactant = 0.00239*129 = 0.308 g

[ Molar mass of CaCl₂.H₂O = 129 g/mol]

3. Total mass of the reactants = 1.00 g

Then, total mass of excess reactant

= (1.00 - 0.308) = 0.692 g

4. Percent limiting reactant

= [ mass of limiting reactant)*100]÷ total mass of reactants

= [0.308)*100]/1.00

= 30.8 % .

5. Percent of excess reactant = ( mass of excess reactant *100)/total mass of reactant

= (0.692*100)/1.00

= 69.2 % .

6. Balanced reaction

K₂C₂O₄.H₂O + CaCl₂.H₂O $\rightarrow$ CaC₂O₄.H₂O + 2KCl + H₂O.

Then , moles of CaC₂O₄.H₂O formed = moles of K₂C₂O₄. H₂O

= 0.00239

Then ,mass of K₂C₂O₄.H₂O (reacted ) = mole * molar mass = 0.00239*184.24 = 0.44 g

[Molar mass of K₂C₂O₄, H₂O = 184.24 g/mol]

7.

Again, mass of excess reactant ( unreacted) = ( 0.692 - 0.44) = 0.252 g