In a reaction vessel, 500. mL of 0.100 M Na3PO4(aq) is combined with 200. mL of 0.400 M Pb(NO3)2(aq).

(a) Write the molecular equation for the reaction that occurs. Balance the equation using lowest whole numbers and be sure to include states.
(b) Identify the limiting reactant. Support your answer with calculations.
(c) Calculate the mass, in grams, of the precipitate that forms.

(a) The molecular equation for the reaction that occurs can be written as follows:

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

(b) To identify the limiting reactant, we need to compare the amount of moles of each reactant.

First, let's find the moles of Na3PO4:
moles of Na3PO4 = volume (L) × concentration (mol/L)
moles of Na3PO4 = 0.500 L × 0.100 mol/L
moles of Na3PO4 = 0.050 mol

Next, let's find the moles of Pb(NO3)2:
moles of Pb(NO3)2 = volume (L) × concentration (mol/L)
moles of Pb(NO3)2 = 0.200 L × 0.400 mol/L
moles of Pb(NO3)2 = 0.080 mol

Now, let's compare the moles of each reactant. The stoichiometry of the balanced equation tells us that the ratio of Na3PO4 to Pb(NO3)2 is 3:2. Therefore, if there is an excess of one reactant, it will be the one with a greater number of moles.

From the calculations, we can see that Na3PO4 has 0.050 mol, while Pb(NO3)2 has only 0.080 mol. Since Pb(NO3)2 has a greater number of moles, it will be the limiting reactant.

(c) To calculate the mass of the precipitate that forms, we need to use the mole ratio between Pb3(PO4)2 and Pb(NO3)2 from the balanced equation.

The molar mass of Pb3(PO4)2 is:
3 × molar mass of Pb + 2 × molar mass of PO4
= 3(207.2 g/mol) + 2(94.97 g/mol)
= 621.6 g/mol + 189.94 g/mol
= 811.54 g/mol

Using the mole ratio from the balanced equation, we can calculate the moles of Pb3(PO4)2 formed from the moles of Pb(NO3)2 reacted:

moles of Pb3(PO4)2 = moles of Pb(NO3)2 × (1 mol Pb3(PO4)2 / 2 mol Pb(NO3)2)
moles of Pb3(PO4)2 = 0.080 mol × (1 mol / 2 mol)
moles of Pb3(PO4)2 = 0.040 mol

Finally, we can calculate the mass of the precipitate formed using the moles of Pb3(PO4)2 and its molar mass:

mass = moles × molar mass
mass = 0.040 mol × 811.54 g/mol
mass ≈ 32.46 g

Therefore, the mass of the precipitate that forms is approximately 32.46 grams.

To answer parts (a) and (b), we need to write the balanced chemical equation for the reaction and determine the limiting reactant.

(a) To write the molecular equation, we need to consider the reaction between sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2). The balanced molecular equation is determined by combining the cations and anions to form the products.

Na3PO4(aq) + Pb(NO3)2(aq) → NaNO3(aq) + Pb3(PO4)2(s)

In this reaction, sodium phosphate reacts with lead(II) nitrate to produce sodium nitrate and lead(II) phosphate, which is a precipitate (s) since it is insoluble in water.

(b) To determine the limiting reactant, we need to compare the moles of each reactant available. First, let's convert the volumes of the solutions into moles using their respective molarities:

For Na3PO4(aq):
Volume = 500 mL = 0.500 L
Molarity = 0.100 M
Moles of Na3PO4 = Molarity × Volume = 0.100 mol/L × 0.500 L = 0.050 mol

For Pb(NO3)2(aq):
Volume = 200 mL = 0.200 L
Molarity = 0.400 M
Moles of Pb(NO3)2 = Molarity × Volume = 0.400 mol/L × 0.200 L = 0.080 mol

Now, we need to compare the mole ratio of Na3PO4 to Pb(NO3)2, which is 1:1. Since the mole ratio is the same for both compounds, the one with fewer moles is the limiting reactant. In this case, Na3PO4 has fewer moles (0.050 mol) compared to Pb(NO3)2 (0.080 mol). Therefore, Na3PO4 is the limiting reactant.

To answer (c), we need to use the limiting reactant to calculate the mass of the precipitate formed.

The stoichiometry of the balanced equation tells us that 1 mole of Na3PO4 reacts with 1 mole of Pb3(PO4)2, so the molar ratio is 1:1. Therefore, the moles of Pb3(PO4)2 formed will be equal to the moles of Na3PO4 used.

Moles of Pb3(PO4)2 = 0.050 mol (from the limiting reactant calculation)

Next, we need to calculate the molar mass of Pb3(PO4)2 to convert moles to grams:

Molar mass of Pb3(PO4)2 = (3 × atomic mass of Pb) + (2 × atomic mass of P) + (8 × atomic mass of O)
= (3 × 207.2 g/mol) + (2 × 31.0 g/mol) + (8 × 16.0 g/mol)
= 621.6 g/mol + 62.0 g/mol + 128.0 g/mol
= 811.6 g/mol

Mass of Pb3(PO4)2 formed = Moles of Pb3(PO4)2 × Molar mass of Pb3(PO4)2
= 0.050 mol × 811.6 g/mol
= 40.58 g

Therefore, the mass of the precipitate that forms is 40.58 grams.