ment 8 Report Sheet Limiting Reactant Lab sec. MIw Name Veronica Radi Date Hao f
ID: 492335 • Letter: M
Question
ment 8 Report Sheet Limiting Reactant Lab sec. MIw Name Veronica Radi Date Hao from the Salt Mixture Precipitation of Caco Trial I Trial 2 1. Mass beaker (s) of 2. Mass of beaker and salt mixture (e) Mass of salt mixture (s) 4, Mass of filter paper (g) 5. Mass of filter paper and product after air-dried OSS or 6. Mass of dried product 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) Data Analysis Hotorcado precipitated (mo) L Caco 2. Moles of limiting reactant in mixture (mo) salt formula of limiting hydrate 3, Mass of limiting reactant in salt mixture (g) Mass of excess reactant in salt mixture formula of excess hydrate 5. Percent limiting reactant in salt mixture 6. Percent excess reactant in salt mixture (9) Mass of excess reactant that reacted (g & Mass of excess reactant, unreacted (g) *Show calculations for Trial 1 on next page. Experiment 8 an orgaExplanation / Answer
trail 2
mass of dreid product = 8.06- 0.822 = 7.238
moles = 0.0.565(mass/ mo.wt of CaC2O4 )
MOL . wt of calcium chloride = 111
MOL . wt of potassium oxalate= 166
total mass of salt mixture as per the balanced equation would be =277 gr.
CaCl2 + K2C2O4-------------> CaC2O4(s) + 2 KCl
salt mixtute taken = 2.06 grams.
weight of calcium chloride = (2.06/277)x 111 = 0.825 grams
weight ofpotassium oxalate = (2.06/277)x 166 =1.235grams( calculated as per stoichiometry) otherwise not given in the question
formula of limiting reactant in the salt mixture CaCl2. H2O
formula of excess reactant in the salt mixture K2C2O4. H2O
generally numbe rof moles = mass / mol .wt
percemt of limiting reagent = 0.825/2.06) x 100 = 40
percemt of lexcess reagent = 1.235/2.06) x 100 = 60
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