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?

Question

Alexis Phelps

Chemistry

9 November, 16:34

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?

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Nakita · Answer 1

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