8.76g of aluminum sulphide is added to 0.0350L of 0.250mol/L lead(II)nitrate solution. calculate the maximum mass of precipitate, PbS that can form
Well, let's do some chemistry calculations while keeping it light and funny!
First, we need to find the limiting reactant. It's like trying to fit a square clown wig on a round clown head - there's a limit to how much can react!
To find the limiting reactant, we compare the number of moles of aluminum sulfide (Al₂S₃) and lead(II) nitrate (Pb(NO₃)₂). Whichever one we have less of is the limiting reactant. It's like having a pie-eating contest with only one pie left!
To find the number of moles of Al₂S₃, we divide the mass (8.76g) by the molar mass (you can look up the exact value). And don't worry, I won't make any "Alumin-yum" jokes here!
Once we have the number of moles of Al₂S₃, we use the balanced chemical equation to determine the molar ratio between Al₂S₃ and PbS. Remember, it's like finding the perfect balance between juggling balls and spinning plates!
Once we find the number of moles of PbS formed, we multiply that by the molar mass of PbS to find the maximum mass of the precipitate. It's like figuring out how much weight you've gained after devouring the entire clown buffet!
So put on your chemistry clown nose and give it a try! Let me know if you need any further assistance along the way.
To calculate the maximum mass of precipitate, PbS, that can form, we need to determine the limiting reactant in the reaction between aluminum sulphide (Al2S3) and lead(II) nitrate (Pb(NO3)2).
First, let's write the balanced chemical equation for the reaction:
3Pb(NO3)2 + 2Al2S3 → 6PbS + 2Al(NO3)3
From the balanced equation, we can see that 2 moles of Al2S3 (molar mass: 2(26.98 g/mol) + 3(32.07 g/mol) = 150.15 g/mol) react with 6 moles of Pb(NO3)2 (molar mass: 207.2 g/mol) to form 6 moles of PbS (molar mass: 6(207.2 g/mol) = 1243.2 g/mol).
Now, let's calculate the moles of Al2S3 added to the solution:
Mass of Al2S3 = 8.76 g
Moles of Al2S3 = (8.76 g) / (150.15 g/mol)
Moles of Al2S3 = 0.0583 mol
Next, let's calculate the moles of Pb(NO3)2 in the solution:
Volume of Pb(NO3)2 solution = 0.0350 L
Molarity of Pb(NO3)2 = 0.250 mol/L
Moles of Pb(NO3)2 = (0.0350 L) * (0.250 mol/L)
Moles of Pb(NO3)2 = 0.00875 mol
From the balanced equation, we can see that the stoichiometric ratio between Al2S3 and Pb(NO3)2 is 2:6, or 1:3.
Therefore, if 0.0583 moles of Al2S3 are reacted, we would require 3 times as much Pb(NO3)2 to completely react with all the Al2S3.
Given that we have 0.00875 moles of Pb(NO3)2, it is clear that Pb(NO3)2 is the limiting reactant because 0.00875 moles is less than the required 0.0583 moles.
Now, we can calculate the maximum mass of PbS that can form. Since the stoichiometric ratio between PbS and Pb(NO3)2 is 6:3, or 2:1, we can calculate the mass of PbS as follows:
Moles of PbS = 0.00875 mol * (6/3)
Moles of PbS = 0.0175 mol
Mass of PbS = Moles of PbS * Molar mass of PbS
Mass of PbS = 0.0175 mol * 1243.2 g/mol
Mass of PbS = 21.7 g
Therefore, the maximum mass of precipitate, PbS, that can form is 21.7 g.
To calculate the maximum mass of precipitate, PbS, that can form in the reaction between aluminum sulphide (Al₂S₃) and lead(II) nitrate (Pb(NO₃)₂), we need to determine the limiting reagent.
1. Write the balanced equation for the reaction:
Al₂S₃ + 3Pb(NO₃)₂ → 2Al(NO₃)₃ + 3PbS
2. Calculate the number of moles of Al₂S₃:
Mass of Al₂S₃ = 8.76 g
Molar mass of Al₂S₃ = 2(26.98 g/mol) + 3(32.06 g/mol) = 150.16 g/mol
Moles of Al₂S₃ = Mass of Al₂S₃ / Molar mass of Al₂S₃
3. Calculate the number of moles of Pb(NO₃)₂:
Volume of Pb(NO₃)₂ solution = 0.0350 L
Concentration of Pb(NO₃)₂ solution = 0.250 mol/L
Moles of Pb(NO₃)₂ = Volume of solution × Concentration
4. Determine the limiting reagent:
The balanced equation shows that the molar ratio of Al₂S₃ to Pb(NO₃)₂ is 1:3.
Compare the moles of Al₂S₃ and Pb(NO₃)₂ to determine which is in excess. The smaller value is the limiting reagent.
5. Calculate the maximum moles of PbS formed:
Since the limiting reagent is Al₂S₃, the maximum moles of PbS formed can be determined from the ratio in the balanced equation:
Moles of PbS = Moles of Al₂S₃ × (3 moles of PbS / 1 mole of Al₂S₃)
6. Calculate the maximum mass of PbS:
Molar mass of PbS = 207.2 g/mol
Mass of PbS = Moles of PbS × Molar mass of PbS
Follow these steps to obtain the maximum mass of precipitate, PbS, that can form in the given reaction involving 8.76 g of Al₂S₃ and 0.0350 L of 0.250 mol/L Pb(NO₃)₂ solution.
Al2S3 + 3Pb(NO3)2 ==>3PbS + 2Al(NO3)3
Convert 8.76 g Al2S3 to moles. moles = grams/molar mass
Convert the Pb(NO3)2 to mols. moles = M x L.
Now, using the coefficients in the balanced equation, convert each of the reactants to moles of PbS. Undoubtedly you will obtain different answers for moles PbS; the correct answer in limiting reagent problems like this is the smaller value. Then convert the smaller value to grams. g = moles x molar mass.