2KI+Pb(NO3)2=2KNO3+PbI2
suppose a solution containing 1.25g of KI is combined with a solution containing 2.42g Pb(NO3)2
_what mass of the yellow PbI would result
what is the limiting reactant
How much in grams of the excess reactant was used
how much in grams of the excess reactant will be left over
This is a long problem. I'm going to give you an example that will work all limiting reagent problems. Just follow the steps.
http://www.jiskha.com/science/chemistry/limiting_reagent_problem.html
To find the mass of PbI2 produced, we need to determine the limiting reactant first.
1. Calculate the number of moles for each reactant:
- Moles of KI: 1.25g / molar mass of KI
- Moles of Pb(NO3)2: 2.42g / molar mass of Pb(NO3)2
The molar mass of KI is found by considering the atomic masses of potassium (K) and iodine (I) and adding them up:
- Mass of K = atomic mass of K = 39.10 g/mol
- Mass of I = atomic mass of I = 126.90 g/mol
- Molar mass of KI = Mass of K + Mass of I
The molar mass of Pb(NO3)2 can be found similarly, considering the atomic masses of lead (Pb), nitrogen (N), and oxygen (O).
2. Determine the stoichiometric ratio of the balanced equation:
From the balanced equation: 2 moles of KI react with 1 mole of Pb(NO3)2 to produce 2 moles of KNO3 and 1 mole of PbI2.
3. Use the stoichiometric ratio to find the limiting reactant:
- Calculate the number of moles of PbI2 that can be produced from 1.25g of KI, using the stoichiometric ratio and mole calculations.
- Calculate the number of moles of PbI2 that can be produced from 2.42g of Pb(NO3)2, using the stoichiometric ratio and mole calculations.
- The reactant that produces the smaller number of moles of PbI2 is the limiting reactant.
4. Calculate the mass of PbI2 produced:
- Use the limiting reactant to calculate the number of moles of PbI2 produced.
- Convert the number of moles to grams by multiplying by the molar mass of PbI2.
5. Determine the mass of the excess reactant used:
- Use the number of moles of the excess reactant and the molar mass to calculate the mass.
6. Calculate the mass of the excess reactant remaining:
- Determine the number of moles of the limiting reactant used using the stoichiometric ratio.
- Subtract the moles of the limiting reactant used from the total moles of the excess reactant.
- Multiply the remaining moles by the molar mass to get the mass.
By following these steps, you will be able to find the mass of the yellow PbI2 produced, identify the limiting reactant, determine the mass of the excess reactant used, and calculate the mass of the excess reactant remaining.
To determine the mass of PbI2 formed, we need to find the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the maximum amount of product that can be formed.
Step 1: Calculate the molar mass of KI and Pb(NO3)2.
- Molar mass of KI: K (39.10 g/mol) + I (126.90 g/mol) = 166.00 g/mol
- Molar mass of Pb(NO3)2: Pb (207.20 g/mol) + 2 * N (14.01 g/mol) + 6 * O (16 + 16 = 32.00 g/mol) = 331.20 g/mol
Step 2: Convert the given masses of KI and Pb(NO3)2 to moles.
- Moles of KI = mass / molar mass = 1.25 g / 166.00 g/mol ≈ 0.00753 mol
- Moles of Pb(NO3)2 = mass / molar mass = 2.42 g / 331.20 g/mol ≈ 0.00731 mol
Step 3: Determine the stoichiometric ratio between KI and Pb(NO3)2.
From the balanced equation, the stoichiometric ratio between KI and Pb(NO3)2 is 2:1.
Step 4: Calculate the moles of PbI2 that can be formed from each reactant.
- Moles of PbI2 formed from KI = 0.00753 mol * (2 mol PbI2 / 2 mol KI) = 0.00753 mol
- Moles of PbI2 formed from Pb(NO3)2 = 0.00731 mol * (1 mol PbI2 / 1 mol Pb(NO3)2) = 0.00731 mol
Step 5: Determine the limiting reactant.
The limiting reactant is the one that produces the least amount of product. In this case, Pb(NO3)2 produces less PbI2, so it is the limiting reactant.
Step 6: Calculate the mass of PbI2 formed from the limiting reactant.
- Mass of PbI2 formed from Pb(NO3)2 = moles * molar mass = 0.00731 mol * 461.01 g/mol ≈ 3.37 g
Therefore, the mass of yellow PbI2 that would result is approximately 3.37 grams.
To calculate the mass of the excess reactant used, we need to find the moles of KI used and subtract it from the total moles of KI initially present.
Step 7: Calculate the moles of KI used.
- Moles of KI used = 0.00753 mol - 0.00731 mol = 0.00022 mol
Step 8: Calculate the mass of the excess reactant used.
- Mass of the excess reactant used = moles * molar mass = 0.00022 mol * 166.00 g/mol ≈ 0.03652 g
Therefore, the mass of the excess reactant (KI) used is approximately 0.03652 grams.
To find the mass of the excess reactant left over, subtract the mass used from the total mass of the excess reactant.
Step 9: Calculate the mass of the excess reactant left over.
- Mass of the excess reactant left over = total mass of excess reactant - mass of excess reactant used
Since the excess reactant is KI (1.25 g):
- Mass of the excess reactant left over = 1.25 g - 0.03652 g ≈ 1.21348 g
Therefore, the mass of the excess reactant (KI) left over is approximately 1.21348 grams.