A student precipitates 397.1 g lead (II) iodide (Pbl₂) in the reaction below. Upon careful calculation, the student determines obtaining a 84.0% yield. What was the theoretical yield of grams of lead (II) iodide (Pbl2) that should have been used? Pb(NO3)²(aq) + 2Kl(aq) → Pblz(s) + 2KNO³(aq) percent yield actual yield/ theoretical yield x 100

Chemistry for Engineering Students
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Chapter3: Molecules, Moles, And Chemical Equations
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Problem 3.14PAE: 3.14 A number of compounds are used in cement, and reactions among them occur when water is added....
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A student precipitates 397.1 g lead (11) iodide (Pbl2) in the reaction below. Upon careful calculation, the student determines obtaining a 84.0% yield. What was
the theoretical yield of grams of lead (11) iodide (Pbl2) that should have been used?
Pb(NO3)²(aq) +2KI (aq) → Pbl2(s) + 2KNO³(aq)
percent yield = actual yield/ theoretical yield x 100
Transcribed Image Text:A student precipitates 397.1 g lead (11) iodide (Pbl2) in the reaction below. Upon careful calculation, the student determines obtaining a 84.0% yield. What was the theoretical yield of grams of lead (11) iodide (Pbl2) that should have been used? Pb(NO3)²(aq) +2KI (aq) → Pbl2(s) + 2KNO³(aq) percent yield = actual yield/ theoretical yield x 100
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