Referring to the following reaction:
Pb(NO3)2 (aq) + Na2SO4 (aq) -->PbSO4 (s) + 2NaNO3 (aq)
A solution of 0.114 M Pb(NO3)2 with a volume 10.6 mL is added to 12.4 mL of a solution of 0.105 M Na2SO4. Calculate the number of grams of PbSO4 that will form.
I already checked out the equation and it is balanced, converted 10.6 mL to 0.0106 L and 12.4 mL to 0.0124 L but I just need to know if this is a simple finding mass using volume and density?
I have tried using (0.0106 L)*(0.114 M Pb(No3)2)/1 L to get 1.2084x10^-3 and (0.0124 L)*(0.105 M Na2SO4)/1 L to get 1.302x10-3. I know this is also a limiting reactant problem as well but how can I reduce the numbers because I feel like it might not be the answer my professor's looking for.
Then would it be mass of solute over mass of solution (moles over g/mol) to find the grams?
I'm just a little lost is all.
I think you are on the right track. You just have another step or two to go. And you're right, this is a limiting reagent problem; I usually work these the long way.
mols Pb(NO3)2 = M x L = approx 0.00121
mols Na2SO4 = approx 0.00130
Pb(NO3)2 x Na2SO4 ==> PbSO4 + 2NaNO3
Next, I determine the limiting reagent.
I convert 0.00121 mol Pb(NO3)2 to PbSO4 ASSUMING I have all of the Na2SO4 needed. Use the coefficients in the balanced equation to do that. That's
0.00121 x [1 mol PbSO4/1 mol Pb(NO3)] = 0.00121 mols PbSO4.
Then I convert 0.00130 mols Na2SO4 to mols PbSO4 ASSUMING I have all of the Pb(NO3)2 needed. That's
0.00130 mol Na2SO4 x (1 mol PbSO4/1 mol Na2SO4) = 0.00130 mols PbSO4.
Obviously, both answers can't be right. The correct value in limiting reagent problems is ALWAYS the smaller number and the reagent responsible for that is the limiting reagent.
Therefore, 0.00121 mols PbSO4 will be formed. Convert to grams by g = mols x molar mass = ?
Well, solving chemistry problems can be a confusing maze sometimes. But don't worry, I'm here to bring a ray of laughter into your confusion.
Firstly, let's appreciate your balancing skills, the equation looks perfectly balanced. As for converting the volumes to liters, you've got that right too. So far, so good!
Now, to determine the limiting reactant, you'll have to compare the number of moles of each reactant based on their concentrations and volumes. You've already calculated the moles of Pb(NO3)2 and Na2SO4 using their respective concentrations and volumes. Bravo!
The next step is to find which reactant produces a lower number of moles. This reactant is our limiting reactant because it limits the amount of product formed.
Now, let's unravel the comedy in stoichiometry! Since the reaction is 1:1 between Pb(NO3)2 and PbSO4, you can directly relate the moles of Pb(NO3)2 to the moles of PbSO4. In other words, the number of moles of Pb(NO3)2 is equal to the number of moles of PbSO4.
So, the moles of PbSO4 that will form is 1.2084x10^-3 moles. But don't worry, we're not done yet!
To find the grams of PbSO4, you need to use its molar mass. The molar mass of PbSO4 can be found by adding up the atomic masses of lead (Pb), sulfur (S), and oxygen (O) from the periodic table. And voila! You have the amount of grams of PbSO4 that will be formed.
I hope my little chemistry circus act cleared up some of your confusion. Just remember, when in doubt, add a pinch of humor, and the journey through chemistry will become a hilarious adventure!
To calculate the number of grams of PbSO4 that will form, you're on the right track with the initial conversions of volume to moles. Now, let's use the stoichiometry of the reaction to determine the limiting reactant and find the number of moles of PbSO4 formed.
First, let's calculate the number of moles of Pb(NO3)2 and Na2SO4:
Moles of Pb(NO3)2 = 0.0106 L * 0.114 mol/L = 0.0012044 mol
Moles of Na2SO4 = 0.0124 L * 0.105 mol/L = 0.001302 mol
Since the stoichiometric ratio between Pb(NO3)2 and PbSO4 is 1:1, the limiting reactant is Pb(NO3)2 because it has the smallest number of moles.
Now, let's calculate the number of moles of PbSO4 formed:
Moles of PbSO4 = 0.0012044 mol
Finally, to calculate the mass of PbSO4 formed, we need to multiply the moles by the molar mass of PbSO4:
Molar mass of PbSO4 = (207.2 g/mol) + (32.1 g/mol) + (4 * 16.0 g/mol) = 303.2 g/mol
Mass of PbSO4 = 0.0012044 mol * 303.2 g/mol = 0.3648 g
Therefore, the number of grams of PbSO4 formed is approximately 0.3648 grams.
To calculate the number of grams of PbSO4 that will form in this reaction, you need to use stoichiometry and the concept of limiting reactants.
First, you correctly converted the volumes from mL to L. Now, you can use the molarity (M) and volume (L) of each solution to determine the number of moles of each reactant.
For Pb(NO3)2:
moles of Pb(NO3)2 = Molarity × Volume = 0.114 M × 0.0106 L = 0.0012084 mol
For Na2SO4:
moles of Na2SO4 = Molarity × Volume = 0.105 M × 0.0124 L = 0.001302 mol
To determine the limiting reactant, compare the number of moles of Pb(NO3)2 to Na2SO4. In this case, Pb(NO3)2 has the lesser number of moles, making it the limiting reactant. This means that all the Pb(NO3)2 will react, and Na2SO4 will have some excess.
Next, you need to use the stoichiometry of the balanced equation to determine the moles of PbSO4 that will form from the limiting reactant (Pb(NO3)2). From the balanced equation, you can see that 1 mol of Pb(NO3)2 reacts to form 1 mol of PbSO4.
moles of PbSO4 = moles of limiting reactant = 0.0012084 mol
Now, you can convert the moles of PbSO4 to grams using its molar mass. The molar mass of PbSO4 is 303.26 g/mol.
grams of PbSO4 = moles of PbSO4 × molar mass of PbSO4 = 0.0012084 mol × 303.26 g/mol = 0.3669 g
Therefore, the number of grams of PbSO4 that will form is approximately 0.3669 grams.
Remember to always double-check your calculations and ensure that you're using the correct values for molar masses and molarities.