Question

* volume of unknown sample (mL) 40.0 mL + mass of filter paper (9) 0.3159 monday mass of filter paper + solid Ba3(PO4)2(g) 49.2939 glassi subtract mass filter paper mass of solid Ba3(PO4)2 (g) from Ba P) a to gethasst Calculation: 1239-0.815:0.1949 0.924g Periodictable molar mass of Ba3(PO4)2 (g/mol) Calculation: 001.92.3782 g/mol comm o go to sus Calculation: m moles of Ba3(PO4)2 mass onday m s 1.4 601.923727 monday 1.535 xioma moles of BaCl2 aquatia 3 to! ratio. / Calculation: 4.605x 10-3 mol nud x 1.535x 103 Stoichiometry Lab

Answer 1

Hi everything seems to be correct, but initial mass of solid Ba₃(PO₄)₂ is incorrect. Please observe below calculations and answers for better clarification as you have not asked me any questions what to do. Main problem comes from g of BaCl₂ per mL of sample value = 0.239 g/mL this is very very high.

Just copy below values to get good credit points or change according to it.

FXPERIMENTAL PROCEDURE 1. Using a 50.0 ml graduated cylinder ng a 300 ml graduated cylinder, hain 40.0 ml of the unknown water sample. This pe contains an unknown amount of Bach. Pour this into a 100 ml beaker. 2. Add 200 ml of the Na PO, solution provided to your unknown solu ution provided to your unknown solution in the 100 ml beaker. The amount of Na PO, you are adding in the 200 ml. of solution is more than enough to provide for complete reaction of the unknown Bach. Consequently, you can be assured 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. 5. Set up the vacuum filtration apparatus. Be sure to clamp the flask to your ring stand. 6. Weigh a piece of filter paper on the balance and record this weight on your data sheet. Stoichiometry Lab

e filter paper in the Buchner funnel. w the filter 4 Connect and turn on the valve on the vacuum line. Using your wash bottle wet the filter paper in the funnel. Wetting the filter paper is to ensure that it is held firmly in the bottom of the funnel; otherwise some of your solid may be pulled under the filter paper and into the suction flask. 9. With the vacuum on, pour your sample into the Buchner funnel. Use a wash bottle to rinse out all of the solid 10. Leave the vacuum running for 5 minutes to aid in the drying of your sample. Turn off the vacuum. 11. Gently remove the filter paper with the sample on it from the Buchner funnel and place it on a watch glass. 12. Label your sample and leave it to dry until the next lab period. 13. Weigh your sample and record the weight on the data sheet. Stoichiometry Lab

Lab Partners): Data Sheet omo volume of unknown sample (m) Back U mass of filter paper (9) 0.339 1.470 mass of filter paper + solid Bas(PO4)2(g) mass of solid Ba3(PO4)2 (9) Teszt 1.149 Calculation: filet paper - 10478-0-339 1-14.00 molar mass of Ba3(PO4)2 (g/mol) Calculation: moles of Bas(PO4)2 Calculation: moles of BaCl Calculation: thiometry Lab

molar mass of BaCl2 (g/mol) Calculation mass of Bach orignally present in the unknown sample (9) Calculation: g of BaCl2 per mL of sample (g/mL) Calculation: Actual calculated value of BaCl2 (g/mL) 0.0125 Experimental Error Calculation:

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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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Please see First picture above

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

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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Hope this helped you!

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