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 (1)/(2)* 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 ~ = ~ \frac {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=(n)/(M)\\\\V=(0.83)/(3)\\\\V=0.277~L`