Question

Based on the data, please answer question 3 (a,b, and c). thank you!

CHEM 2115 Limiting Reagent Lab Report #6 Chem I Lab Name Mikhayla D'Anne Section# Station# A5 Date 1%4/19 Overall equation: CaCl (aq) + K:C:04H:0 (aq) = CaC,O4+H20 (8) +2 KCI (aq) Net lonic Equation: Ca"? (aq) + CO. (aq) + H20 (1 = CaC20. H:0 (5) 1. Data & Calculations : CaCl K2C2O4+H2O Concentration of solution 0.75 mol/L Volume of solution 15 ml Moles of CaCl2 used 5.92x104 mol Mass assigned 0.1409 Mass used 0.1429 Moles of K CO-H2O used 0.0008 mai (MW - 184.24 g/mol) Crucible and filter mass CaC,0..H:0 (Precipitate) Dry (crucible + filter) 22.1159 With Precipitate 22,2219 Mass of CaC0H20 recovered 0.1069 Moles of CaCO..H2O recovered 0.0007 mol (MW-146.12 g/mol) Be sure to include the necessary units for the values reported in the above table. Attach sample calculations of each type necessary to obtain the results in the above table. These should include the equations used, substituted values, appropriate units, and results. 2. Excess and Limiting Tests: 21. Did a precipitate form (1) when K C O solution was added to the filtrate? Yes (2) when CaCl solution was added? No 2b. Based on the results shown in 2a, what was the limiting reactant? K 2,04 What was the excess reactant? Cacz

Answer 1

Calzag imole + K₂C 20 4 H₂O aq 1 fo CaC2O4 H₂O6 + 2 kcaq I mole 2 mole I mole - from y 5.82x104 calculation mole 2 01125 mol . 0008 mole 0007 mole moles af cael = conecutoction & Volume .75 mol x 15ml = 1125 X10 1000 mt = .01125- mole Imole of caca real with 1.00 molek,620; H₂O 01125 = .01125 mole : K₂ C₂O4 H₂O is less than calculation it is limiting reactant & cacle is in excess 9ts are of H. 24. form = Imole Cacout20 . .. I mole ka co ou H₂O aq - .oood mole " " = 0oodmole 1 bereent veild Experimental yeild x 100 Theoratical yeild 0007 0008 Xloo 1 x 100 = 87.5%. - Ang