Could you please help solve the following? I have A-C.. When a solution of Lead
ID: 545142 • Letter: C
Question
Could you please help solve the following? I have A-C..
When a solution of Lead Nitrate, Pb(NO3)2 , is mixed with a solution of sodium iodide, NaI, a precipitate of lead iodide, PbI2 , is formed.
a) Write the chemical equation for this reaction showing the state of all reactants and products.
b) If 1 mol of Pb(NO3)2 reacts with 2 mols of NaI, what is the amount of PbI2 that will be formed?
c) If 1 mol of Pb(NO3)2 reacts with 2 mols of NaI, which compound would be the limiting reagent?
d) For the mixture described in c), what is the amount of PbI2 that forms?
e) Solution A contains 50.00 ml of 1.000 x 10-1 mol x L-1 Pb(NO3)2 and solution B contains 50.00 ml of 1.000 x 10-1 mol x L-1 NaI. Calculate the mass of Pb(NO3)2 in solution A and the mass of NaI in solution B.
f) When solution A and B described above are mixed, solid PbI2 forms. Calculate the theoretical yield of PbI2.
g) 1.021 g of solid PbI2 was actually collected from the mixture above. Calculate the percent yield of PbI2.
Explanation / Answer
ANSWERS FOR A-C
a) Balanced chemical equation between Pb(NO3)2 and NaI
Pb(NO3)2 + 2NaI = PbI2 + 2NaNO3
b) Pb(NO3)2 2NaI PbI2
1 mole 2 moles amount formed = 461g
amount of PbI2 formed = atomic weight of Pb (207) + 2 * atomic weight of I (127)
= 207 + 254 = 461 amu
c) The limiting reagent is Pb(NO3)2. as it is completely consumed in the reaction to form one mole of PbI2