Question

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq)+2NH4I(aq)⟶PbI2(s)+2NH4NO3(aq) What volume of a 0.250 M NH4I solution is required to react with 121 mL of a 0.360 M Pb(NO3)2 solution? volume: mL How many moles of PbI2 are formed from this reaction? moles: mol PbI2

Answer 1

Answer:

a) volume of ammonium iodide required =349 mL

b) the moles of lead iodide formed = 0.0436 mol

Explanation:

The reaction is:

$Pb(NO_{3})_{2}+2NH_{4}I -->PbI_{2}+2NH_{4}NO_{3}$

It shows that one mole of lead nitrate will react with two moles of ammonium iodide to give one mole of lead iodide.

Let us calculate the moles of lead nitrate taken in the solution.

Moles=molarityX volume (L)

Moles of lead nitrate = 0.360 X 0.121 =0.0436 mol

the moles of ammonium iodide required = 2 X0.0436 = 0.0872 mol

The volume of ammonium iodide required will be:

$volume=\frac{moles}{molarity}=\frac{0.0872}{0.250}=0.349L=349mL$

the moles of lead iodide formed = moles of lead nitrate taken = 0.0436 mol