20.0 mL of a 0.125 M solution of potassium iodide is added to 15.0 mL of a 0.235 M solution of lead (II) nitrate. A reaction occurs via the balanced equation below:
2KI(aq)+ Pb(NO3)2(aq) --> 2KNO3(aq) +PBI2(ppt)
How much product can theoretically be produced in the reaction described above?
This is a limiting reagent problem. You know that because amounts for BOTH reactants are given.
Process:
1. mols reactant 1 = ?
2. mols reactant 2 = ?
3. calculate mols product using reactant 1 without regard to reactant 2.
4. Calculate mols product using reactant 2 without regard to reactant 1.
5. It is likely that mols product from steps 3 and 4 will not be the same; the correct answer in limiting reagent problems is ALWAYS the smaller value and the reactant producing that value is the limiting reagent.
6. The grams of the product = theoretical yield = mols (the smaller value) x molar mass.
Here is a step by step example if you need it.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
thank you so much!
To determine the amount of product that can theoretically be produced in the given reaction, we need to use stoichiometry and the concept of limiting reagents.
1. First, we need to determine the limiting reagent. The limiting reagent is the reactant that is completely consumed first, thus limiting the amount of product formed.
2. Convert the given volumes of solutions to moles using the molarity of each solution and the volume given.
For potassium iodide (KI):
Molarity (M) = 0.125 M
Volume (V) = 20.0 mL = 0.0200 L
Number of moles (n) = Molarity × Volume = 0.125 M × 0.0200 L = 0.0025 moles
For lead (II) nitrate (Pb(NO3)2):
Molarity (M) = 0.235 M
Volume (V) = 15.0 mL = 0.0150 L
Number of moles (n) = Molarity × Volume = 0.235 M × 0.0150 L = 0.00353 moles
3. Use the balanced equation to determine the stoichiometric ratio between the reactants and the product.
From the balanced equation: 2 moles of KI react with 1 mole of Pb(NO3)2 to produce 1 mole of PBI2.
4. We can now compare the amounts of reactants to the stoichiometric ratio.
The stoichiometric ratio tells us that 2 moles of KI react with 1 mole of Pb(NO3)2 to form 1 mole of PBI2. Therefore, for every 2 moles of KI, we need 1 mole of Pb(NO3)2.
Given that we have 0.0025 moles of KI and 0.00353 moles of Pb(NO3)2, we can conclude that Pb(NO3)2 is the limiting reagent because we have fewer moles of Pb(NO3)2 compared to the stoichiometric ratio.
5. Use the stoichiometric ratio to determine the amount of PBI2 produced.
From the balanced equation: 2 moles of KI react to produce 1 mole of PBI2.
Since Pb(NO3)2 is the limiting reagent, the number of moles of PBI2 that can be produced is equal to half the number of moles of Pb(NO3)2.
Number of moles of PBI2 = 0.00353 moles of Pb(NO3)2 / 2 = 0.001765 moles of PBI2
6. Convert the moles of PBI2 to grams using the molar mass of PBI2.
Molar mass of PBI2 = 390.99 g/mol
Mass of PBI2 = Number of moles of PBI2 × Molar mass of PBI2 = 0.001765 moles × 390.99 g/mol = 0.6881 grams
Therefore, the theoretical yield of PBI2 in the reaction is approximately 0.6881 grams.