Question

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb(NO3)2(aq) + 2NH,I(aq) — Pl(s) + 2NH, NO, (aq) What volume of a 0.270 M NH I solution is required to react with 487 mL of a 0.680 M Pb(NO3)2 solution? volume: How many moles of Pbly are formed from this reaction? moles: mol Pbl2

Answer 1

One mole of lead nitrate forms one mole of lead iodide.

Mvi Mare M = Molarity of NHy I = 0.270M - Volume of why I = ? n - number of moles of Why I = 2 mole M = Molarity of Pb(NO3)2 = 0.680M ₂ - volume of Pb (voz) = 487mL ₂ – No-of moles of Pb (bg) - 1 mole obsoxt M₂ V₂ m, v=Mt M X- 0-680 M X 487mL amore 18 mols -0.270 M - 138 m 2453mL. No. of mole of ph(NO) in 487mL of 0.6804 Pb ) = 0.680x487 _ 0.331 mots . Number of mots of pbs formed = 0.33 ! 1000 -

Molarity is the number of moles of solute present in one litre of solution.

M = number of moles of solute*1000/volume of solution in mL

Number of moles of solute = Molarity of solution*Volume of solution in mL/1000

Number of moles of Lead nitrate present in 487 mL of 0.680 M Pb(NO_{​​​​​3})₂

= 0.680 M * 487 mL/1000

= 0.331 moles

0.331 moles of lead iodide is formed.