What is the theoretical yield of PbI2 with 25.0 g of Pb(NO3)2?
Actually you need a reaction to be certain but in a pinch you can do
25.0 g Pb(NO3)2 x [molar mass PbI2/molar mass Pb(NO3)2] = ??g PbI2.
To calculate the theoretical yield of PbI2, you need to follow these steps:
Step 1: Write out the balanced chemical equation.
The balanced equation for the reaction between Pb(NO3)2 and KI to form PbI2 is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
Step 2: Calculate the molar mass of Pb(NO3)2.
Pb(NO3)2 has one Pb atom weighing 207.2 g/mol, two N atoms weighing 2 x 14.01 g/mol, and six O atoms weighing 6 x 16.0 g/mol. Adding them together gives:
207.2 g/mol + 2(14.01 g/mol) + 6(16.0 g/mol) = 331.2 g/mol
Step 3: Convert the given mass of Pb(NO3)2 to moles.
To do this, divide the given mass by the molar mass:
25.0 g / 331.2 g/mol = 0.0753 mol
Step 4: Determine the stoichiometry of the reaction.
From the balanced equation, you can see that 1 mole of Pb(NO3)2 reacts with 1 mole of PbI2.
Step 5: Calculate the theoretical yield of PbI2.
Since the stoichiometry is 1:1 between Pb(NO3)2 and PbI2, the number of moles of PbI2 produced will be equal to the number of moles of Pb(NO3)2 used. Therefore, the theoretical yield of PbI2 is also 0.0753 mol.
Step 6: Convert the moles of PbI2 to grams.
The molar mass of PbI2 is one Pb atom weighing 207.2 g/mol and two I atoms weighing 2 x 126.9 g/mol. Adding them gives:
207.2 g/mol + 2(126.9 g/mol) = 460.0 g/mol
To convert moles to grams, multiply the number of moles by the molar mass:
0.0753 mol x 460.0 g/mol = 34.6 g
Therefore, the theoretical yield of PbI2 is 34.6 grams when starting with 25.0 grams of Pb(NO3)2.
To find the theoretical yield of PbI2, you need to use the stoichiometry of the balanced chemical equation that relates Pb(NO3)2 to PbI2.
The balanced equation for the reaction is:
Pb(NO3)2 + 2 KI → PbI2 + 2 KNO3
From the balanced equation, you can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to produce 1 mole of PbI2.
Here's how you can calculate the theoretical yield of PbI2:
1. Convert the given mass of Pb(NO3)2 to moles.
The molar mass of Pb(NO3)2 is 331.2 g/mol.
(25.0 g Pb(NO3)2) / (331.2 g/mol) = 0.0755 moles Pb(NO3)2
2. Use the mole ratio from the balanced equation to determine the moles of PbI2 that can be formed.
From the balanced equation, the mole ratio of Pb(NO3)2 to PbI2 is 1:1.
Therefore, the moles of PbI2 formed will also be 0.0755 moles.
3. Convert the moles of PbI2 to grams.
The molar mass of PbI2 is 461.0 g/mol.
(0.0755 moles PbI2) * (461.0 g/mol) = 34.8 grams PbI2 (rounded to one decimal place)
Therefore, the theoretical yield of PbI2 is approximately 34.8 grams when starting with 25.0 grams of Pb(NO3)2.