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 0.590 M NH4I solution is required to react with 225 mL of a 0.680 M Pb(NO3)2 solution?

Answer 1

Answer:

0.519 L of a 0.590M NH4I solution is required

Explanation:

Step 1: The balanced equation

Pb(NO3)2(aq) + 2NH4I(aq) → PbI2(s) + 2NH4NO3(aq)

Step 2: Given data

Molarity of NH4I = 0.590 M

Volume of Pb(NO3)2 = 225 mL = 0.225 L

Molarity of Pb(NO3)2 = 0.680 M

Step 3: Calculate number of moles Pb(NO3)2

Number of moles = Molarity * volume

Number of moles Pb(NO3)2 = 0.680 M * 0.225L

Number of moles Pb(NO3)2 = 0.153 moles

Step 4: Calculate moles of NH4I

For 1 mol Pb(NO3)2 consumed, we need 2 moles NH4I

For 0.153 moles of Pb(NO3)2, we have 0.306 moles of NH4I

Step 6: Calculate volumen of NH4I

Molarity = moles of NH4I / volume

0.590 M = 0.306 moles / volumen

volume =0.306 moles / 0.590M

volume = 0.519 L

0.519 L of a 0.590M NH4I solution is required