Question

Name: Lab Partner(s): Data Sheet volume of unknown sample (mL) Bad UD-om mass of filter paper (g) mass of filter paper + solid Ba3(PO4)2(g) 0.339 1.479 Calculation: mass of solid Ba3(PO4)2 (9) filter paper -- Total 1647g-0-339 = 1.149 11.149 molar mass of Baz(PO4)2 (a/mol)

Answer 1

given mass of solid Ba₃(PO₄)₂ = 1.14 g

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molar mass of Ba₃(PO₄)₂ = [ ( 3 x atomic mass of Ba) + [(2 x [( atomic mass of P + (4 x atomic mass of O)]]]

molar mass of Ba₃(PO₄)₂ = [ (3 x 137.327 ) + [ 2 x [( 30.973762) + ( 4 x 15.9994 )]]] = 601.9237 g/mol

molar mass of Ba₃(PO₄)₂ = 601.9237 g/mol

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moles of Ba₃(PO₄)₂ = mass of Ba₃(PO₄)₂ / molar mass of Ba₃(PO₄)₂ = 1.14 g / 601.9237 g/mol = 0.0019 mol

moles of Ba₃(PO₄)₂ = 0.0019 mol

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moles of BaCl₂

balanced chemical reaction equation is: 3 BaCl₂ (aq) + 2 Na₃PO₄ (aq) ==> Ba₃(PO₄)₂ (s) + 6 NaCl (aq)

3 mol BaCl₂ gives 1 mol Ba₃(PO₄)₂ , so 0.0019 mol of Ba₃(PO₄)₂ are formed from 3 x 0.0019 mol = 0.0057 mol BaCl₂

moles of BaCl₂ = 0.0057 mol

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mass of BaCl₂ originally present in the unknown solution = moles of BaCl₂ x molar mass of BaCl₂

mass of BaCl₂ originally present in the unknown solution = 0.0057 mol BaCl₂ x 208.2330 g/mol = 1.1869281 g

mass of BaCl₂ originally present in the unknown solution = 1.19 g

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Unknown water sample actually is 40.0 mL, even we measured in 50.0 mL graduated cylinder & transferred it into 100 mL beaker. But addition of 20 mL of Na₃PO₄ volume to be considered, so total volume = 40 + 20 = 60 mL

EXPERIMENTAL PROCEDURE 1. Using a 50.0 mL graduated cylinder, obtain 40.0 mL of the unknown water sample. This ample contains an unknown amount of BaCh, Pour this into a 100 mL beaker 2. Add 20.0 ml. of the NasPO solution provided to your unknown solution in the 100 ml, beaker The amount of NaPOs you are adding in the 20.0 ml, of solution is more than enough to provide for complete reaction of the unknown BaClh, Consequently, you can be assured that your reaction has gone to completion. 3. Stir your mixture with a stirring rod, off and on for 2-3 minutes to give time for complete reaction. 4. Let your reaction mixture settle for 10 minutes while you set up the filtration apparatus. The precipitate should settle to the bottom making filtration casier. Set up the vacuum filtration apparatus. Be sure to clamp the flask to your ring stand. 5. Weigh a piece of filter paper on the balance and record this weight on your data sheet. 6. Stoichiometry Lab

g of BaCl₂ per mL = 1.19 g / 60.0 mL = 0.01983333333 g/mL

g of BaCl₂ per mL = 0.0198 g / mL

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Actual calculated value of BaCl₂ (g/mL) given as = 0.0125 g/mL

Experimental error = [|experimental - actual| / actual] x 100% = ([0.0198 - 0.0125] / 0.0125) x 100% = 58.4 %

Experimental error = 58.4 %

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