Calculation of the moles of the products.

Question

the precipitation reaction of PB(NO3)ii and KI solution.

The book asked us to calculate in moles of each reactant and product of when the PB(NO3)ii will be 0.0035moles (this

solution is 3.5 cm3 with the concentration of 1.0 mole/dm-3) ; and KI solution will be 0.005moles (this solution is 5 cm3 with

the concentration of same above) reacts, what is the amount in moles of the products?

The book answers are n(PbI2) = 0.0035 mol, n(KNO3) = 0.00175 mol.

But according to my calculation, the precipitation i.e. PbI2 should be 0.0025 moles (because KI is a limiting reactant) and the

KNO3 will be 0.005 moles. Which answer is correct?

Answer 1

The balanced equation will be

Pb(NO3)2 + 2KI ---> 2KNO3 + PbI2

3.5 mmol 5.0 mmol ? ?

Note 1 dm3 = 1 L, 1 cm3 = 1 mL, 1 mmol = 10-3 mol

Clearly, the most PbI2 that will be produced will be half of 5.0 mmol, since the coefficients are in ratio 2:1. There is no way that the 3.5 mmol of Pb nitrate will completely react to give 3.5 mmol.

So you are absolutely right. IMHO