Consider the following reaction:
3Pb(NO3)2(aq) +Al2(SO4)3(aq) ¡÷ 3PbSO4(aq) + 2Al(NO3)3(aq)
Determine the mass (in gram) of lead(II) sulphate (PbSO4) and aluminium nitrate [Al(NO3)3] formed in the solution when 8.84 g of lead(II) nitrate [Pb(NO3)2] was reacted with 6.28 g of aluminium shlphate [Al2(SO4)3].
This is a limiting reagent (LR) problem and you know that because amounts for BOTH reactants is given.
mols Pb(NO3)2 = grams/molar mass = ?
mols Al2(SO4)3 = grams/molar mass = ?
Using the coefficients in the balanced equation, convert mols Pb(NO3)2 to mols either product.
Do the same and convert mols Al2(SO4)3 to mols of the SAME product.
It is likely these values will not agree which means one of them is not right; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that value is the LR. It is essential at this step to identify the LR. Using the smaller value of the product, convert to grams. g = mols x mlar mass.
For the other product, use the LR and convert with the coefficients to determine mols of the other product and convert to grams as above.
To determine the mass of lead(II) sulfate (PbSO4) and aluminum nitrate (Al(NO3)3) formed, we need to calculate the moles of the reactants and use the stoichiometric coefficients from the balanced equation.
Step 1: Calculate the moles of lead(II) nitrate (Pb(NO3)2).
Given: mass of Pb(NO3)2 = 8.84 g
Molar mass of Pb(NO3)2 = 331.2 g/mol
Moles of Pb(NO3)2 = mass / molar mass = 8.84 g / 331.2 g/mol = 0.0267 mol
Step 2: Calculate the moles of aluminum sulfate (Al2(SO4)3).
Given: mass of Al2(SO4)3 = 6.28 g
Molar mass of Al2(SO4)3 = 342.2 g/mol
Moles of Al2(SO4)3 = mass / molar mass = 6.28 g / 342.2 g/mol = 0.0183 mol
Step 3: Determine the limiting reactant.
To determine the limiting reactant, we compare the moles of each reactant to their stoichiometric coefficients.
From the balanced equation:
3Pb(NO3)2(aq) + Al2(SO4)3(aq) -> 3PbSO4(aq) + 2Al(NO3)3(aq)
The stoichiometric ratio between Pb(NO3)2 and PbSO4 is 3:3 (1:1), and between Al2(SO4)3 and PbSO4 is 1:3. This means that for every 3 moles of Pb(NO3)2 reacted, we should expect 3 moles of PbSO4.
The stoichiometric ratio between Al2(SO4)3 and PbSO4 is 1:3. This means that for every 1 mole of Al2(SO4)3 reacted, we should expect 3 moles of PbSO4.
Comparing the moles of Pb(NO3)2 and Al2(SO4)3:
Moles of Pb(NO3)2 = 0.0267 mol
Moles of Al2(SO4)3 = 0.0183 mol
Since 0.0183 mol < 0.0267 mol, Al2(SO4)3 is the limiting reactant.
Step 4: Calculate the moles of PbSO4 formed.
From the balanced equation, the stoichiometric ratio between Al2(SO4)3 and PbSO4 is 1:3.
Moles of PbSO4 = moles of Al2(SO4)3 * (3 moles of PbSO4 / 1 mole of Al2(SO4)3) = 0.0183 mol * (3/1) = 0.0549 mol
Step 5: Calculate the mass of PbSO4 formed.
Molar mass of PbSO4 = 303.3 g/mol
Mass of PbSO4 = moles of PbSO4 * molar mass = 0.0549 mol * 303.3 g/mol = 16.67 g
Therefore, the mass of lead(II) sulfate (PbSO4) formed is 16.67 grams.
Step 6: Calculate the moles of Al(NO3)3 formed.
From the balanced equation, the stoichiometric ratio between Al2(SO4)3 and Al(NO3)3 is 2:1.
Moles of Al(NO3)3 = moles of Al2(SO4)3 * (2 moles of Al(NO3)3 / 1 mole of Al2(SO4)3) = 0.0183 mol * (2/1) = 0.0366 mol
Step 7: Calculate the mass of Al(NO3)3 formed.
Molar mass of Al(NO3)3 = 213.0 g/mol
Mass of Al(NO3)3 = moles of Al(NO3)3 * molar mass = 0.0366 mol * 213.0 g/mol = 7.79 g
Therefore, the mass of aluminum nitrate (Al(NO3)3) formed is 7.79 grams.
To determine the mass of lead(II) sulfate (PbSO4) and aluminum nitrate (Al(NO3)3) formed in the solution, we first need to calculate the number of moles of the reactants and then use stoichiometry to find the mole ratio and mass of the products.
1. Calculate the number of moles of lead(II) nitrate (Pb(NO3)2):
- Given mass of Pb(NO3)2 = 8.84 g
- Determine the molar mass of Pb(NO3)2:
- Pb: atomic mass = 207.2 g/mol
- N: atomic mass = 14.01 g/mol
- O: atomic mass = 16.00 g/mol (3 oxygen atoms in each nitrate)
- Therefore, molar mass of Pb(NO3)2 = (207.2 g/mol) + (2 * (14.01 g/mol)) + (6 * (16.00 g/mol)) = 331.2 g/mol
- Calculate the number of moles of Pb(NO3)2 by dividing the given mass by the molar mass:
- Number of moles = Mass / Molar mass = 8.84 g / 331.2 g/mol ≈ 0.0267 mol
2. Calculate the number of moles of aluminum sulfate (Al2(SO4)3):
- Given mass of Al2(SO4)3 = 6.28 g
- Determine the molar mass of Al2(SO4)3:
- Al: atomic mass = 26.98 g/mol
- S: atomic mass = 32.06 g/mol
- O: atomic mass = 16.00 g/mol (4 oxygen atoms in each sulfate)
- Therefore, molar mass of Al2(SO4)3 = (2 * (26.98 g/mol)) + (3 * (32.06 g/mol) + (12 * (16.00 g/mol)) = 342.2 g/mol
- Calculate the number of moles of Al2(SO4)3 by dividing the given mass by the molar mass:
- Number of moles = Mass / Molar mass = 6.28 g / 342.2 g/mol ≈ 0.0184 mol
3. Determine the mole ratio between Pb(NO3)2 and PbSO4:
- From the balanced equation, we can see that the mole ratio between Pb(NO3)2 and PbSO4 is 3:3 or 1:1.
4. Determine the mole ratio between Al2(SO4)3 and Al(NO3)3:
- From the balanced equation, we can see that the mole ratio between Al2(SO4)3 and Al(NO3)3 is 2:2 or 1:1.
5. Calculate the mass of PbSO4 formed:
- We know that the number of moles of PbSO4 is equal to the number of moles of Pb(NO3)2 (0.0267 mol) due to the 1:1 mole ratio.
- Determine the molar mass of PbSO4:
- Pb: atomic mass = 207.2 g/mol
- S: atomic mass = 32.06 g/mol
- O: atomic mass = 16.00 g/mol (4 oxygen atoms in each sulfate)
- Therefore, molar mass of PbSO4 = (207.2 g/mol) + (32.06 g/mol) + (4 * (16.00 g/mol)) = 399.2 g/mol
- Calculate the mass of PbSO4 by multiplying the number of moles by the molar mass:
- Mass = Number of moles * Molar mass = 0.0267 mol * 399.2 g/mol ≈ 10.66 g
6. Calculate the mass of Al(NO3)3 formed:
- We know that the number of moles of Al(NO3)3 is equal to the number of moles of Al2(SO4)3 (0.0184 mol) due to the 1:1 mole ratio.
- Determine the molar mass of Al(NO3)3:
- Al: atomic mass = 26.98 g/mol
- N: atomic mass = 14.01 g/mol
- O: atomic mass = 16.00 g/mol (3 oxygen atoms in each nitrate)
- Therefore, molar mass of Al(NO3)3 = (26.98 g/mol) + (3 * (14.01 g/mol)) + (9 * (16.00 g/mol)) = 212.9 g/mol
- Calculate the mass of Al(NO3)3 by multiplying the number of moles by the molar mass:
- Mass = Number of moles * Molar mass = 0.0184 mol * 212.9 g/mol ≈ 3.89 g
Therefore, the mass of lead(II) sulfate (PbSO4) formed in the solution is approximately 10.66 grams, and the mass of aluminum nitrate (Al(NO3)3) formed is approximately 3.89 grams.