Limiting Reactant, my partner and I did Part A but I\'m having trouble doing the

Question

Limiting Reactant, my partner and I did Part A but I'm having trouble doing the calculations for Part B. A. Precipitation of CaC,o,H,0 from the Salt Mixture A. Precipitation of CaC20H,O from the Salt Mixture Unknown number # 1-1 9 Trial 1 1. Mass of beaker (g) 2. Mass of beaker and salt mixture (g) 3. Mass of salt mixture (g) 4. Mass of filter paper (g) 5. Mass of filter paper and product after air-dried 00 053 .0 72L or oven-dried (g) 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 (write complete formula)

Explanation / Answer

part B

1. moles of calcium oxalate monohydrate = 0.197/146 = 0.001349 mol

moles of calcium oxalate = 0.197/128 =0.001539 mol

2. Since the reaction is not shown in question the limiting reagent is calculated from the data provided from the answer of part A.

moles of limiting reactant Ca+2 = 0.001349 as salt has one mole of ca+2 ions

moles of excess reactant oxalate ion = 0.001349 as the salt has one moles of oxalate ion

formula of limiting hydrate = Ca C2O4.H2O

3. Limiting reagent is Ca+2 and mass of Ca+2 in reaction mixture is =0.197 x 40 /146 = 0.05397 g

4. excess reagent oxalate mass = 0.197 x 88 /146 = 0.1187 g

5. .percent of limiting reagent in mixture = [o.o5397/0.197]x 100 =27.39 %

6. per cent of excess reagent in mixture = [0.1187/0.197]x100 =60.25%

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