When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed lead sulfate precipitates out of soultion. Calculate the mass of lead (II) sulfate that should form when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are reacted together.

Pb(NO3)2+ Na2SO4 ==> PbSO4 + 2NaNO3

moles Pb(NO3)2 = 1.25L x 0.0500M = 0.0625.
moles Na2SO4 = 2L x 0.0250 = 0.0500.
Since 1 mol Pb(NO3)2 uses 1 mol Na2SO4 to form 1 mol PbSO4, mol PbSO4 formed must be 0.0500. There isn't enough Na2SO4 for all of the Pb(NO3)2 to be used.
Convert 0.0500 mols PbSO4 to grams. g= mols x molar mass = ?

Well, well, well, it seems we have a chemistry problem on our hands. Don't worry, I'm here to help and hopefully make you chuckle along the way!

To figure this out, we first need to find out the moles of each compound involved, and then determine the limiting reactant. But let's do this with a little twist of humor, shall we?

Step 1: Calculate the moles of Pb(NO3)2

We have 1.25 L of a 0.0500 M solution. That sounds like a lot of chemistry mumbo-jumbo, but trust me, it's simpler than it sounds. So, 1.25 L times 0.0500 M gives us... *drumroll*... 0.0625 moles of Pb(NO3)2. We're getting closer!

Step 2: Calculate the moles of Na2SO4

Now, we move on to the 2.00 L of a 0.0250 M solution. Just like before, we multiply these two values together: 2.00 L times 0.0250 M, which gives us... another *drumroll*... 0.0500 moles of Na2SO4. Time for the big finale!

Step 3: Determine the limiting reactant

Since we have more moles of Pb(NO3)2 (0.0625 moles) compared to Na2SO4 (0.0500 moles), the limiting reactant is Na2SO4. It's like a chemistry version of "Two men enter, one man leaves!" Sorry, I couldn't resist a Mad Max reference.

Step 4: Calculate the mass of PbSO4 formed

Now that we know our limiting reactant, we can calculate the mass of PbSO4. We know that 1 mole of Pb(NO3)2 produces 1 mole of PbSO4. So, the molar mass of PbSO4 is 303.26 g/mol (I hope you have a calculator handy because I'm not doing the math here).

Finally, multiplying the moles of PbSO4 by its molar mass (0.0500 moles x 303.26 g/mol) should give us the mass of PbSO4 that should form! Gotta love chemistry, right? It's like doing magic tricks with formulas.

So, my dear friend, plug in those numbers and let the potion of PbSO4 reveal its mass to you!

To calculate the mass of lead (II) sulfate that forms when the two solutions are mixed, we need to follow these steps:

Step 1: Write down the balanced chemical equation for the reaction:

Pb(NO3)2 + Na2SO4 -> PbSO4 + 2 NaNO3

Step 2: Calculate the number of moles of each reactant:

Number of moles of Pb(NO3)2 = volume (L) × concentration (M)
Number of moles of Pb(NO3)2 = 1.25 L × 0.0500 M = 0.0625 mol

Number of moles of Na2SO4 = volume (L) × concentration (M)
Number of moles of Na2SO4 = 2.00 L × 0.0250 M = 0.0500 mol

Step 3: Determine the limiting reactant:

The limiting reactant is the reactant that is completely consumed in the reaction and determines the amount of product formed. To find the limiting reactant, compare the number of moles of each reactant using the stoichiometric ratio from the balanced equation.

From the balanced equation, we can see that the ratio of Pb(NO3)2 to PbSO4 is 1:1. Therefore, 0.0625 mol of Pb(NO3)2 will react to form 0.0625 mol of PbSO4.

From the balanced equation, we can see that the ratio of Na2SO4 to PbSO4 is 1:1. Therefore, 0.0500 mol of Na2SO4 will react to form 0.0500 mol of PbSO4.

Comparing the moles of the two reactants, it is clear that Na2SO4 is the limiting reactant, as it will be completely consumed.

Step 4: Calculate the mass of PbSO4 formed:

Now that we know that Na2SO4 is the limiting reactant, we can use its mole amount to calculate the moles of PbSO4 formed.

Number of moles of PbSO4 = 0.0500 mol

The molar mass of PbSO4 can be calculated using the atomic masses of lead (Pb), sulfur (S), and oxygen (O):

Molar mass of PbSO4 = (1 × atomic mass of Pb) + (1 × atomic mass of S) + (4 × atomic mass of O)
Molar mass of PbSO4 = (1 × 207.2 g/mol) + (1 × 32.1 g/mol) + (4 × 16.0 g/mol)
Molar mass of PbSO4 = 303.3 g/mol

Mass of PbSO4 formed = number of moles × molar mass
Mass of PbSO4 formed = 0.0500 mol × 303.3 g/mol = 15.165 g

Therefore, the mass of lead (II) sulfate that should form when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are reacted together is 15.165 grams.

15.1633 g PbSO4

15.2g

15.2