60 mL of 0.322 M potassium iodide are combined with 20.0 mL of 0.530 m lead(II) nitrate. How many grams of lead (II) iodide will precipitate?
To determine the amount of lead (II) iodide that will precipitate, we need to find the limiting reactant first. This is the reactant that will be completely consumed and determine the maximum amount of product that can be formed.
Step 1: Calculate the number of moles of potassium iodide (KI):
Given:
Volume of KI solution = 60 mL = 0.060 L
Concentration of KI = 0.322 M
Number of moles (n) = concentration (C) × volume (V)
n(KI) = 0.322 M × 0.060 L = 0.01932 moles of KI
Step 2: Calculate the number of moles of lead (II) nitrate (Pb(NO3)2):
Given:
Volume of Pb(NO3)2 solution = 20.0 mL = 0.0200 L
Concentration of Pb(NO3)2 = 0.530 M
n(Pb(NO3)2) = 0.530 M × 0.0200 L = 0.0106 moles of Pb(NO3)2
Step 3: Use the balanced chemical equation to find the stoichiometric ratio of KI to PbI2:
2 KI + Pb(NO3)2 → 2 KNO3 + PbI2
From the equation, we see that 2 moles of KI react with 1 mole of Pb(NO3)2 to produce 1 mole of PbI2.
Step 4: Determine the limiting reactant:
Comparing the number of moles of KI (0.01932 moles) to Pb(NO3)2 (0.0106 moles), we see that Pb(NO3)2 is the limiting reactant since it has fewer moles.
Step 5: Calculate the moles of PbI2 formed:
From the stoichiometric ratio, 1 mole of Pb(NO3)2 reacts to produce 1 mole of PbI2. Therefore, the moles of PbI2 formed will be equal to the moles of Pb(NO3)2 used.
n(PbI2) = 0.0106 moles
Step 6: Calculate the mass of PbI2 formed:
To calculate the mass, we need to use the molar mass of PbI2.
Molar mass of PbI2 = atomic mass of Pb + 2 × atomic mass of I
= (207.2 g/mol) + 2 × (126.9 g/mol)
= 207.2 g/mol + 253.8 g/mol
= 461.0 g/mol
Mass PbI2 = moles of PbI2 × molar mass of PbI2
= 0.0106 moles × 461.0 g/mol
Therefore, the mass of lead (II) iodide that will precipitate is 4.87 grams (rounded to two decimal places).
To find the number of grams of lead(II) iodide that will precipitate, we need to determine the limiting reactant first. The limiting reactant is the one that is completely consumed first, and it determines the maximum amount of product that can be formed.
1. Convert the given volumes of solutions into moles:
- For potassium iodide (KI):
- Volume: 60 mL
- Concentration: 0.322 M
- Molar mass of KI: 39.10 g/mol + 126.90 g/mol = 166.00 g/mol (atomic masses of K and I)
- Moles of KI = Volume (L) × Concentration (M) = 0.060 L × 0.322 mol/L = 0.01932 mol
- For lead(II) nitrate (Pb(NO3)2):
- Volume: 20.0 mL
- Concentration: 0.530 M
- Molar mass of Pb(NO3)2: 207.20 g/mol + 2 × (14.01 g/mol + 16.00 g/mol) + 6 × (16.00 g/mol + 3 × (16.00 g/mol)) = 331.20 g/mol
- Moles of Pb(NO3)2 = Volume (L) × Concentration (M) = 0.0200 L × 0.530 mol/L = 0.0106 mol
2. Find the stoichiometric ratio between the reactants and the product. The balanced chemical equation for the reaction between lead(II) nitrate and potassium iodide is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
From the equation, we see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2.
3. Determine the limiting reactant:
To do this, we need to compare the moles of both reactants. Since the stoichiometric ratio is 1:2 for Pb(NO3)2:KI, we need twice as many moles of KI as moles of Pb(NO3)2 to react completely.
Moles of Pb(NO3)2 = 0.0106 mol
Moles of KI needed = 2 × Moles of Pb(NO3)2 = 2 × 0.0106 mol = 0.0212 mol
The moles of KI available (0.01932 mol) are less than moles of KI needed (0.0212 mol). Hence, KI is the limiting reactant.
4. Calculate the moles of lead(II) iodide (PbI2) formed:
From the stoichiometry, we know that 1 mole of PbI2 is formed for every mole of KI reacted.
Moles of PbI2 formed = Moles of KI = 0.01932 mol
5. Calculate the mass of PbI2 formed:
- Molar mass of PbI2: 207.20 g/mol + 2 × 126.90 g/mol = 460.00 g/mol
- Mass of PbI2 formed = Moles of PbI2 formed × Molar mass of PbI2 = 0.01932 mol × 460.00 g/mol ≈ 8.88 g
Therefore, approximately 8.88 grams of lead(II) iodide (PbI2) will precipitate.
2KI + Pb(NO3)2 = PbI2 + 2KNO3
You will get half as many moles of PbI2 as you use moles of KI
so now just see which gets used up first, and that will tell you how many grams are produced.