Answer:
105 grams PbI₂
Explanation:
Pb(NO₃)₂ + 2KI => 2KNO₃ + PbI₂(s)
moles Pb(NO₃)₂ = 0.265L(1.2M) = 0.318 mole
moles KI = 0.293(1.55M) = 0.454 mole => Limiting Reactant
moles PbI₂ from mole KI in excess Pb(NO₃)₂ = 1/2(0.454 mole) = 0.227 mol PbI₂
grams PbI₂ = 0.227 mol PbI₂ x 461 g/mole = 104.68 g ≈ 105 g PbI₂(s)
Answer:
104.6 grams of PbI2 (solid) will be produced
Explanation:
Step 1: Data given
Volume of a 1.20 M lead(II) nitrate = 265.0 mL = 0.265 L
Volume of a 1.55 M potassium iodide = 293.0 mL = 0.293 L
Step 2: The balanced equation
Pb(NO3)2(aq) + 2KI(aq) → 2KNO3(aq) + PbI2(s)
Step 3: Calculate moles
Moles = molarity * volume
Moles Pb(NO3)2 = 1.20 M * 0.265 L
Moles Pb(NO3)2 = 0.318 moles
Moles KI = 1.55 M * 0.293 L
Moles KI = 0.454 moles
Step 4: Calculate the limiting reactant
For 1 mol Pg(NO3)2 we need 2 moles of KI to produce 2 moles of KNO3 and 1 mol of PbI2
Ki is the limiting reactant. It will completely be consumed. (0.454 moles) Pb(NO3)2 is in excess. There will react 0.454 / 2 = 0.227 moles
There will remain 0.318 moles - 0.227 =0.091 moles
Step 5: Calculete moles PbI2
For 1 mol Pg(NO3)2 we need 2 moles of KI to produce 2 moles of KNO3 and 1 mol of PbI2
For 0.454 moles KI we'll have 0.454/2 = 0.227 moles PbI2
Step 6: Calculate mass PbI2
Mass PbI2 = moles PbI2 * molar mass PbI2
Mass PbI2 = 0.227 moles * 461.01 g/mol
Mass PbI2 = 104.6 grams
104.6 grams of PbI2 (solid) will be produced
