Answer:
The mass of PbI2 will be 18.2 grams
Explanation:
Step 1: Data given
Volume solution = 99.8 mL = 0.0998 L
mass % KI = 12.0 %
Density = 1.093 g/mL
Volume of the other solution = 96.7 mL = 0.967 L
mass % of Pb(NO3)2 = 14.0 %
Density = 1.134 g/mL
Step 2: The balanced equation
Pb(NO3)2(aq) + 2 KI(aq) ⇆ PbI2(s) + 2 KNO3(aq)
Step 3: Calculate mass
Mass = density * volume
Mass KI solution = 1.093 g/mL * 99.8 mL
Mass KI solution = 109.08 grams
Mass KI solution = 109.08 grams *0.12 = 13.09 grams
Mass of Pb(NO3)2 solution = 1.134 g/mL * 96.7 mL
Mass of Pb(NO3)2 solution = 109.66 grams
Mass of Pb(NO3)2 solution = 109.66 grams * 0.14 = 15.35 grams
Step 4: Calculate moles
Moles = mass / molar mass
Moles KI = 13.09 grams / 166.0 g/mol
Moles KI = 0.0789 moles
Moles Pb(NO3)2 = 15.35 grams / 331.2 g/mol
Moles Pb(NO3)2 = 0.0463 moles
Step 5: Calculate the limiting reactant
For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3
Ki is the limiting reactant. It will completely be consumed ( 0.0789 moles). Pb(NO3)2 is in excess. There will react 0.0789/2 = 0.03945 moles. There will remain 0.0463 - 0.03945 = 0.00685 moles
Step 6: Calculate moles PbI2
For 1 mol Pb(NO3)2 we need 2 moles KI to produce 1 mol PbI2 and 2 moles KNO3
For 0.0789 moles KI we'll have 0.0789/2 = 0.03945 moles PbI2
Step 7: Calculate mass of PbI2
Mass PbI2 = moles PbI2 * molar mass PbI2
Mass PbI2 = 0.03945 moles * 461.01 g/mol
Mass PbI2 = 18.2 grams
