A solution containing excess lead(ii) nitrate is reacted with 380.0ML of 0.250mol/L potassium iodide solution. A bright yellow Precipitate is of lead(ii) iodide is formed.
Calculate the mass of lead (ii) iodide that should be formed.
Pb(NO3)2+2KI-> 2KNO3+PbI2
mols KI = M x L = ?
Use the coefficients in the balanced equation to convert mols KI to mols PbI2.
Now convert mols PbI2 to grams. g = mols x molar mass = ?
To calculate the mass of lead (II) iodide (PbI2) formed, we need to use stoichiometry. Stoichiometry involves using the balanced chemical equation to relate the amounts of reactants and products.
Given:
- Volume of potassium iodide (KI) solution = 380.0 mL
- Concentration of potassium iodide (KI) solution = 0.250 mol/L
Step 1: Convert the volume of the solution to moles of KI using its concentration.
Moles of KI = Volume of KI solution (L) × Concentration of KI solution (mol/L)
Since the volume is given in mL, we need to convert it to liters:
Volume of KI solution (L) = 380.0 mL ÷ 1000 mL/L
Volume of KI solution (L) = 0.380 L
Now, we can calculate the moles of KI:
Moles of KI = 0.380 L × 0.250 mol/L
Moles of KI = 0.095 mol
Step 2: Use the stoichiometry of the balanced chemical equation to relate the moles of KI to moles of PbI2.
From the balanced equation: 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form 1 mole of PbI2.
Since the stoichiometry is 1:1 for KI to PbI2, the moles of KI will also be the moles of PbI2 formed.
Moles of PbI2 formed = 0.095 mol
Step 3: Calculate the molar mass of PbI2.
Molar mass of PbI2 = 207.2 g/mol (atomic masses: Pb = 207.2 g/mol, I = 126.9 g/mol)
Step 4: Convert moles of PbI2 to grams using molar mass.
Mass of PbI2 = Moles of PbI2 × Molar mass of PbI2
Mass of PbI2 = 0.095 mol × 207.2 g/mol
Mass of PbI2 = 19.684 g
Therefore, the mass of lead (II) iodide (PbI2) that should be formed is approximately 19.684 grams.