Question

Properties Chemical List: lodine anion l'(lon) (3.163941 g) (0.024932 moles) Nitrate ion NO3- (lon) (1.872129 g) (0.030193 moles) Unknown (lon) (0.920241 g) Water H20 (Liquid) (80.000000 9) (4.440744 moles) -200 150 Molecule Viewer 2501100 ml L 50 Sample Weight: 5.044 g Chemical Reaction: OK
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1. How many grams of lead (II) iodide would be produced based on the amount of lead (II) nitrate used?

2. What is the theoretical yield of lead (II) iodide (in grams) for this reaction based on the limiting reactant?

3. What was the actual yield (in grams) of the solid for this experiment?

4. What was the percent yield of the reaction?

5. How many moles of iodide were left unreacted?

6. How many moles of NaI were left unreacted?

7. How many grams of sodium iodide were left reacted?

8. How many grams of NaI reacted?

Answer 1

Molu if I = 0.024932 ~ Culoas Molu of No2 = 0.030193 ~ (0.0302 Molu of H₂O ~ u. yo - Mas if lead (11) lodida (Pb Iz) = 5.044 goam Moler man of PbI 2 = 461.01 g/mol o » Moles of Pb Iz = 5.044 0.01094 moles 461.0) Pb (103) + 2 Nai Pb Izt mal 2 NaNO3 required Pb(NO3)₂ + 2 Na] > PbIzt 2 Na Nos reaction 0.030) 10.025 (Mols of No2 = 0.0302 Moles 20151) initially = 60151 of Pb (103)2 = 0.0302 o 0 0.025 0.0151 NaI il the Note: 0.025 < 0.0151, limiting reagent

-> p b Izt 2 NaNO3 Pb(NO3)₂ + 2 Nas 0-01S 0-0250 0 O initially tinally 0 ools-00125 = 0.0026 0.0125 0.025 (1). Mola of Pb Iz the critically, 0.0125 moles Mall of Pol, the was to cally = Molux Moles man = 0-0125 X 461.0 - 5.76 26 (11) Actual yield=5.044 gram (iii) % yield - Actual yield .100 5.044 xloo The writical yield 5-7626 goans Х yo yielde 87.53%) 11 Molu ol NaI left unreacted 0 Molis of I left unreacted = 0 (vi) Molu NaI reacted= 0.025, Moles mais of Nal = 109.89,-. Mau = 0.025 X 149.89 ul 3.74725 goom (vii) Mol. Maus of Na] duft starter = 3.74725 gram