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)
<u>Step 2:</u> 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
<u>Step 3</u>: 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
<u>Step 4:</u> 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
<u>Step 6:</u> 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