The balanced equation:
Pb(NO3)2 + 2KI
PbI2(S) + 2KNO3
After calculating the number of moles of each reactant, the KI is the limiting reagent. Thus not
all Pb(NO3)2 will react.
Formed PbI2 is solid so it doesn’t dissolve into Pb and I ions, so we will calculate the Pb ions in
the excess Pb(NO3)2 (the quantity that doesn’t react = number of moles of Pb(NO3)2 – number
of moles of reacted Pb(NO3)2). Then divide the number of moles over the total volume to get
the concentration.
Solution:
a)
- N of Pb(NO3)2 = 0.33 moles / It can form 0.33 moles of PbI2
- N of KI = 0.555 / It can form 0.2775 moles of PbI2
KI is the limiting reagent.
Then 0.2775 moles of Pb(NO3)2 will react with the 0.555 of KI.
The unreacted Pb(NO3)2 = 0.33 – 0.2775 =0.0525 moles
N of moles of Pb ions = 0.0525 moles
0.0525
Its concentration = (275+300)×10−3 = 0.0913 𝑀
A is true
b) Mass of PbI2= number of moles x molar mass
= 0.2775 x 461 = 128 g
B is true
0.555
c) Concentration of K ions = (275+300)×10−3 = 0.965 M
0.33
d) Concentration of NO3 ions = (275+300)×10−3 = 0.574 𝑀
D is false