Lead(II) carbonate decomposes to form Lead(II) oxide and carbon dioxide. If 5.oo grams of the reactant decomposes to produce 1.52 grams of Lead(II) oxide experimentally, what is the theoretical yield and percent yield.
PbCO3 ---> PbO + CO2
First, calculate theoretical yield:
5.00g*(1 mole/267.21 g) = moles of PbCO3
moles of PbCO3=moles of PbO
moles of PbO*(223.20 g/mole)= mass of PbO
mass of PbO=theoretical yield
(1.52g/theoretical yield)*100= % Yield
when heated lead nitrate decomposes to lead oxide, nitrogen dioxide and oxygen gas
To find the theoretical yield and percent yield, we need to perform some calculations.
1. Calculate the molar mass of lead(II) carbonate (PbCO3):
- The molar mass of Pb = 207.2 g/mol
- The molar mass of C = 12.01 g/mol
- The molar mass of O = 16.00 g/mol
- Therefore, the molar mass of PbCO3 = (1 * 207.2) + (1 * 12.01) + (3 * 16.00) = 267.21 g/mol
2. Calculate the number of moles of lead(II) carbonate (PbCO3):
- Moles = Mass / Molar mass
- Moles = 5.00 g / 267.21 g/mol = 0.0187 mol
3. Based on the balanced chemical equation, the molar ratio between lead(II) carbonate (PbCO3) and lead(II) oxide (PbO) is 1:1.
- This means that for every 1 mole of lead(II) carbonate, we should obtain 1 mole of lead(II) oxide.
4. Calculate the theoretical yield of lead(II) oxide (PbO):
- Theoretical yield = Moles of PbCO3 * Molar mass of PbO
- The molar mass of PbO = 207.2 g/mol
- Theoretical yield = 0.0187 mol * 207.2 g/mol = 3.88 g
5. Calculate the percent yield:
- Percent yield = (Actual yield / Theoretical yield) * 100
- Percent yield = (1.52 g / 3.88 g) * 100 = 39.18%
Therefore, the theoretical yield of lead(II) oxide is 3.88 grams, and the percent yield is 39.18%.
To find the theoretical yield and percent yield, we first need to calculate the stoichiometric ratio between the reactant and the product. The balanced equation for the decomposition of lead(II) carbonate is:
PbCO3 → PbO + CO2
From the equation, we can see that the molar ratio between PbCO3 and PbO is 1:1. This means that for every mole of lead(II) carbonate (PbCO3), we will produce one mole of lead(II) oxide (PbO).
Now, let's calculate the number of moles of PbCO3 and PbO:
1. Calculate the number of moles of PbO produced:
- We know the experimental mass of PbO is 1.52 grams.
- First, determine the molar mass of PbO. The molar mass of Pb is 207.2 g/mol, and the molar mass of O is 16.0 g/mol.
- So, the molar mass of PbO is 207.2 + 16.0 = 223.2 g/mol.
- Now, we can calculate the number of moles of PbO:
Moles of PbO = Mass of PbO / Molar mass of PbO = 1.52 g / 223.2 g/mol = 0.00681 mol
2. Since the balanced equation states that the molar ratio between PbCO3 and PbO is 1:1, the number of moles of PbCO3 is also 0.00681 mol.
3. The theoretical yield is the maximum amount of product that can be formed under ideal conditions. In this case, it is the number of moles of PbO, which is 0.00681 mol.
4. Now, let's calculate the percent yield:
- The percent yield is the actual yield (experimental yield) divided by the theoretical yield, multiplied by 100.
- We are given that the experimental yield of PbO is 1.52 grams.
- First, convert the experimental yield from grams to moles using the molar mass of PbO:
Moles of PbO (experimental yield) = Mass of PbO (experimental yield) / Molar mass of PbO = 1.52 g / 223.2 g/mol = 0.00681 mol
- Percent yield = (Moles of PbO (experimental yield) / Moles of PbO (theoretical yield)) * 100
Percent yield = (0.00681 mol / 0.00681 mol) * 100 = 100%
Therefore, the theoretical yield of PbO is 0.00681 mol, and the percent yield is 100%.