What volume of a 0.130 M NH4I solution is required to react with 905 mL of a 0.280 M Pb (NO3) 2 solution?

3.90 L

Explanation:

The reaction between NH₄I and Pb (NO₃) ₂ is a double replacement reaction, so, the ions will dissociate and change in the two substances. The ions are NH₄⁺, I⁻, Pb⁺², and NO₃⁻, so the reaction is:

2NH₄I + Pb (NO₃) ₂ → PbI₂ + 2NH₄NO₃

Thus, by the stoichiometry of the reactions, 2 moles of NH₄I are necessary to react with 1 mol of Pb (NO₃) ₂. According to Proust's law, the proportion of the reaction must be kept so:

2 moles/1 mol = n NH₄I/n Pb (NO₃) ₂

The number of moles of Pb (NO₃) ₂ that will react is the concentration multiplied by the volume in L, so:

n Pb (NO₃) ₂ = 0.280 * 0.905 = 0.2534 mol

2/1 = n NH₄I/0.2534

n NH₄I = 0.5068 mol

The volume of NH₄I is:

n NH₄I = 0.130 * V

0.5068 = 0.130V

V = 3.90 L