79 LAB 8 STOICHIOMETRY AND LIMITING REACTANTS DATA COLLECTED Unknown number of s

Question

79 LAB 8 STOICHIOMETRY AND LIMITING REACTANTS DATA COLLECTED Unknown number of salt mixture #t05 1. Trial 1 Trial 2 mass of empty vial mass of vial + salt mixture mass of dry filter paper ,759 a. 910g 50p 50/ mass of fiter paper + dry precipitate S0/.29-.73a .50l t.59/ reagent causing additional precipitation Ballb all CALCULATIONS (See example calculations) Trial 1 Trial 2 mass of salt mixture used mass of precipitate formula of limiting reagert formula of excess reagent calculated mass of limiting reagent calculated mass of excess reagent mass % of limiting reagent in sample mass % of excess reagent in sample d 30

Explanation / Answer

The mass of precipitate = 0.281 g

The formula of the precipitate = Ba3(PO4)2

The molar mass of Ba3(PO4)2 = 602 g mol-1

Therefore, the no. of moles of Ba3(PO4)2 = 0.281/602 = 4.668*10-4 mol

Since the BaCl2 reagent is causing the precipitation of Ba3(PO4)2, it should be completely consumed in the reaction, hence BaCl2 should be the limiting reagent. You can cross check the calculation as shown below.

According to the balanced equation, if you take 3 moles of BaCl2, you will get 1 mole of Ba3(PO4)2

i.e. The no. of moles of BaCl2 = 3*4.668*10-4 mol = 0.0014 mol

The molar mass of BaCl2 = 208 g mol-1

Therefore, the mass of BaCl2 = 0.0014 mol * 208 g mol-1 = 0.291 g

The mass of salt mixture = 2.88 g

i.e. The mass of Na3PO4 = 2.88 - 0.291 = 2.589 g

The molar mass of Na3PO4 = 164 g mol-1

Therefore, the no. of moles of Na3PO4 = 2.589/164 = 0.016 mol

According to the balanced equation, for 3 moles of BaCl2, 2 moles of Na3PO4 is required.

But here, for 0.0014 moles of BaCl2, 0.016 moles of Na3PO4 is used.

Hence, BaCl2 is consumed completely and is the limiting reagent.

And Na3PO4 is the excess reagent.

The calculated mass of limiting reagent = 0.291 g

The calculated mass of excess reagent = 2.589 g

The mass% of limiting reagent in the sample = (0.291/2.88)*100 = 10.1%

The mass% of excess reagent in the sample = (2.589/2.88)*100 = 89.9%

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