If 25g of lead phosphate reacts with an excess of sodium nitrate, how many dreams of sodium phosphate is produced?

I can tell you this will never do it because the spontaneous reaction is the reverse of what you have. However, here is how you work the theoretical (and it's ony theoretical) problem.

1. Write and balance the equation.
Pb3(PO4)2 +6NaNO3 ==> 2Na3PO4 + 3Pb(NO3)2

2. Convert grams Pb3(PO4)2 to mols. mols = grams/molar mass.

3. Using the coefficients in the balanced equation, convert mols Pb3(PO4)2 to mols Na3PO4.

4. Convert mols Na3PO4 to grams. g = mols x molar mass. This is the theoretical yield

To find the number of moles of sodium phosphate produced, we can use stoichiometry.

First, we need to balance the chemical equation for the reaction between lead phosphate (Pb3(PO4)2) and sodium nitrate (NaNO3):

3Pb3(PO4)2 + 10NaNO3 → 6Na3PO4 + 3Pb(NO3)2

From the balanced equation, we can see that for every 3 moles of Pb3(PO4)2, we obtain 6 moles of Na3PO4. This means that the molar ratio between Pb3(PO4)2 and Na3PO4 is 3:6, which simplifies to 1:2.

To calculate the number of moles of lead phosphate, we need to divide the given mass of lead phosphate (25g) by its molar mass. The molar mass of Pb3(PO4)2 is:

(3 x atomic mass of Pb) + (2 x atomic mass of P) + (8 x atomic mass of O)

(3 x 207.2 g/mol) + (2 x 31.0 g/mol) + (8 x 16.0 g/mol) = 811.2 g/mol

So, the number of moles of lead phosphate is:

moles = mass ÷ molar mass
moles = 25g ÷ 811.2 g/mol

Now, we can calculate the number of moles of sodium phosphate. Since the molar ratio between Pb3(PO4)2 and Na3PO4 is 1:2, the number of moles of Na3PO4 is also twice the number of moles of Pb3(PO4)2.

moles of Na3PO4 = 2 x moles of Pb3(PO4)2

Finally, to convert the number of moles of sodium phosphate to grams, we can use its molar mass. The molar mass of Na3PO4 is:

(3 x atomic mass of Na) + atomic mass of P + (4 x atomic mass of O)

(3 x 22.99 g/mol) + 30.97 g/mol + (4 x 16.0 g/mol) = 163.97 g/mol

gram = moles x molar mass
gram = moles of Na3PO4 x 163.97 g/mol

Now, you can substitute the calculated values to find the answer.