Calculation for the preparation of the Na2S04 solution: V=100ml x 1 L/1000mL = 0

Question

Calculation for the preparation of the Na2S04 solution: V=100ml x 1 L/1000mL = 0.100 L Mass of Na2SO4-MxFwXY = 0.10 M x 142.04 g/mol x 0.100 L = 1.4205 g =1.4g Actual mass weighed: 1.402 g Actual concentration of solution Exact Mass of Unknown Mixture Used 1. Calculate the mass of the limiting reagent for each trial. My limiting reagent was BaC12 Trial #1 Trial #2 1.008 g 1.000 g Mass of Filter Paper 2. Calculate the number of moles of your limiting reagent from the mass of your precipitate. Trial #1 Trial #2 0.867 g 0.891 g 3. What is the averaged mass percentage of Ba 2+contained in the unknown sample? Mass of Precipitate Trial #1 Trial #2 0.278 0.294

Explanation / Answer

BaCl2 + Na2SO4 ---> BaSO4(ppt) + 2NaCl

Trial#1,

mass of precipitate = 0.278 g

moles of BaSO4 = 0.278 g/233.4 g/mol = 0.0012 mol

moles Na2SO4 taken = 1.402 g/142.04 g/mol = 0.0099 mol

moles of Na2SO4 taken was greater than the moles of BaSO4 formed, so BaCl2 would be the limiting reagent here.

1. mass of BaCl2 (limiting reagent) = 0.0012 mol x 208.23 g/mol = 0.25 g

2. moles of BaCl2 (limiting reactant) = 0.0012 mol

mass% Ba2+ in the sample = 0.0012 mol x 137.33 g/mol x 100/1.008 g = 16.35%

Trial#2,

mass of precipitate = 0.294 g

moles of BaSO4 = 0.294 g/233.4 g/mol = 0.00126 mol

moles Na2SO4 taken = 1.402 g/142.04 g/mol = 0.0099 mol

moles of Na2SO4 taken was greater than the moles of BaSO4 formed, so BaCl2 would be the limiting reagent here.

1. mass of BaCl2 (limiting reagent) = 0.00126 mol x 208.23 g/mol = 0.26 g

2. moles of BaCl2 (limiting reactant) = 0.00126 mol

mass% Ba2+ in the sample = 0.00126 mol x 137.33 g/mol x 100/1.008 g = 17.30%

3. Average mass pecentage of Ba2+ contained in the unknown sample = 16.825%

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