Lead(ll) Iodide can be produced from the reaction of Lead(ll) nitrate and potassium Iodide.What mass of lead(ll) iodide is produced when 50.0 mL of each solution is combined,forming a 0.750 M Solution of potassium nitrate?
Please help me to solve this problem thank you
7 years ago
7 years ago
PUP Senior Highschool <3
11 months ago
To solve this problem, you'll need to follow several steps.
Step 1: Write and balance the chemical equation for the reaction.
The balanced equation for the reaction between lead(II) nitrate (Pb(NO3)2) and potassium iodide (KI) is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
Step 2: Calculate the number of moles of Pb(NO3)2 and KI in the solution.
We know that the concentration (Molarity) is given as 0.750 M. The formula for calculating the number of moles is Moles = Concentration × Volume (in liters).
Given:
Concentration of Pb(NO3)2 = 0.750 M
Volume of Pb(NO3)2 = 50.0 mL = 0.0500 L
Number of moles of Pb(NO3)2 = 0.750 M × 0.0500 L = 0.0375 moles
Similarly, the number of moles of KI can be calculated in the same way.
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the stoichiometry of the balanced equation. In this case, the stoichiometry tells us that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2.
Since the stoichiometric ratio is 1:1, we can conclude that the limiting reactant is Pb(NO3)2 because it will be completely consumed before all the KI.
Step 4: Calculate the mass of PbI2 produced.
To calculate the mass of PbI2 produced, we use the equation:
Mass = Number of moles × Molecular weight
The molecular weight of PbI2 is the sum of the atomic weights of lead (207.2 g/mol) and iodine (126.9 g/mol).
Number of moles of PbI2 = 0.0375 moles
Molecular weight of PbI2 = (207.2 g/mol) + (2 × 126.9 g/mol) = 460.0 g/mol
Mass of PbI2 = 0.0375 moles × 460.0 g/mol = 17.25 grams
Therefore, when 50.0 mL of each solution is combined, 17.25 grams of lead(II) iodide (PbI2) will be produced.