Question
Asked Nov 13, 2019
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A precipitation reaction occurs when 655 mL of 0.639 M Pb(NO2), reacts with 313 mL of 0.855 M KI, as shown by the
following equation.
Pb(NO3)2(aq) 2 KI(aq)Pbl, (s) + 2 KNO3 (aq)
Identify the limiting reactant.
O Pb(NO3)2
O KI
O Pbl2
Ο ΚNO,
KNO3
Calculate the theoretical yield of PbI, from the reaction.
mass of Pbl,:
Calculate the percent yield of PbI, if 48.0 g of Pbl, are formed experimentally.
percent yield of PbI,:
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A precipitation reaction occurs when 655 mL of 0.639 M Pb(NO2), reacts with 313 mL of 0.855 M KI, as shown by the following equation. Pb(NO3)2(aq) 2 KI(aq)Pbl, (s) + 2 KNO3 (aq) Identify the limiting reactant. O Pb(NO3)2 O KI O Pbl2 Ο ΚNO, KNO3 Calculate the theoretical yield of PbI, from the reaction. mass of Pbl,: Calculate the percent yield of PbI, if 48.0 g of Pbl, are formed experimentally. percent yield of PbI,:

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Expert Answer

Step 1

The given volume and molarity of Pb(NO3)2 is 655 mL and 0.639 M.

The given volume and molarity of KI is 313 mL and 0.855 M.

 

The given reaction is as follows:

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Pb(NO,),(aq)+ 2KI(aq)-Pbl, (s)+2KNO, (aq)

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Step 2

It is known that 1 mL = 0.001 L.

 

Therefore, the volume of Pb(NO3)2 in L is 0.655 L.

Similarly, the volume of KI in L is 0.313 L.

 

The mathematical expression of molarity is,

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Number of moles of solute Molarity = Volume of solution in litres

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Step 3

Put the given values in the formula to calculate the nu...

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Number of moles of solute Molarity Volume of solution in litres Number of moles of Pb(NO,) 0.639 0.655 Number of moles of Pb(NO3 ), = 0.4185

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