(Refer to the Procedures for answering this question.) Suppose that 9.58 ml. of
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Question
(Refer to the Procedures for answering this question.) Suppose that 9.58 ml. of the iron(III) chloride solution used in this experiment is added to a beaker containing potassium oxalate hydrate dissolved in water. How many moles of FeCl_3 have been added to the beaker. Refer to the following reaction to answer the questions below: 3Zn(N0_3)_2 6H_2O + 2Na_3PO_4 I2H_20 right arrow Zn_3(PO_4)_2 4H_2O + 6NaNO_3 + 7H_O Determine the limiting reagent if 0.0451 moles of Zn(N0_3)_2.6H_2O is reacted with 0.0337 moles ofNa_3PO_4. 12H_O. Calculate the mass of Zn_3(PO_4)4H_2O product expected to be obtained. Calculate the percent yield if 5.72 g. of Zn(PO_4)_2. 4H_O product was actually recovered in the lab.Explanation / Answer
3 Zn(NO3)2.6H2O + 2 Na3PO4.12H2O ------> Zn3(PO4)2.4H2O + 6 NaNO3 + 7 H2o
3 moles of Zn(NO3)2.6H2O produces 1 mole of Zn3(PO4)2.4H2O
0.0451 moles of Zn(NO3)2.6H2O produces (0.0451/3) moles of Zn3(PO4)2.4H2O
= 0.01503 moles
2 moles of Na3PO4.12H2O produces 1 mole of Zn3(PO4)2.4H2O
0.0337 moles of Na3PO4.12 H2O produces (0.0337/2) noles of Zn3(PO4)2.4H2O
= 0.01685 moles
so limiting reagent is Zn3(PO4)2.12H2O
Number of moles of Zn3(PO4)2.4H2O recovered in the lab = 5.72 / 458.14
= 0.01248 moles
percentage yield = 0.01248 / 0.01503 * 100 = 83.03 %
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