Question

Given the unbalanced equation below, answer the following: Calculate the number of liters of 3.00 M lead (II) iodide solution produced when 1.66 mol Kl react?
Pb(NO3)2 + 2KI → 2KNO3 + PbI2

Answer 1

The number of liters of 3.00 M lead (II) iodide : 0.277 L

Further explanation

Reaction(balanced)

Pb(NO₃)₂(aq) + 2KI(aq) → 2KNO₃(aq) + PbI₂(s)

moles of KI = 1.66

From the equation, mol ratio of KI : PbI₂ = 2 : 1, so mol PbI₂ :

$\tt \dfrac{1}{2}\times 1.66=0.83$

Molarity shows the number of moles of solute in every 1 liter of solute or mmol in each ml of solution

$\large \boxed {\bold {M ~ = ~ \dfrac {n} {V}}}$

Where

M = Molarity

n = Number of moles of solute

V = Volume of solution

So the number of liters(V) of 3.00 M lead (II) iodide-PbI₂ (n=0.83, M=3):

$\tt V=\dfrac{n}{M}\\\\V=\dfrac{0.83}{3}\\\\V=0.277~L$