15.0 mL of 0.30 M sodium phosphate solution reacts with 20.0 mL of 0.23M lead(II) nitrate solution. What mass of precipitate will form?

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To determine the mass of the precipitate that will form, we need to first identify the balanced chemical equation for the reaction between sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2).

The balanced chemical equation for the reaction is:

3Na3PO4 + 2Pb(NO3)2 -> 6NaNO3 + Pb3(PO4)2

From the balanced equation, we can see that 3 moles of sodium phosphate reacts with 2 moles of lead(II) nitrate to form 1 mole of lead(II) phosphate (Pb3(PO4)2).

To determine the number of moles of the reactants, we can use the equation:

moles = concentration (M) x volume (L)

For the sodium phosphate solution:
moles of sodium phosphate = 0.30 M x 0.015 L = 0.0045 moles

For the lead(II) nitrate solution:
moles of lead(II) nitrate = 0.23 M x 0.020 L = 0.0046 moles

Since the balanced equation shows that 3 moles of sodium phosphate reacts with 2 moles of lead(II) nitrate, we can determine the limiting reactant by comparing the moles of the two reactants.

From the calculation above, we see that the number of moles of sodium phosphate (0.0045 moles) is less than the number of moles of lead(II) nitrate (0.0046 moles). Therefore, sodium phosphate is the limiting reactant.

Now, we can calculate the number of moles of the precipitate (lead(II) phosphate) formed using the stoichiometry of the balanced equation.

From the balanced equation:
3 moles of sodium phosphate produce 1 mole of lead(II) phosphate

Therefore, the number of moles of lead(II) phosphate formed is:
moles of lead(II) phosphate = (0.0045 moles of sodium phosphate) / 3 = 0.0015 moles

To determine the mass of the precipitate, we use the molar mass of lead(II) phosphate, which can be calculated using the atomic masses of lead (Pb) and phosphorus (P).

The molar mass of lead(II) phosphate (Pb3(PO4)2) is:
molar mass = (207.2 g/mol x 3) + (30.97 g/mol x 2) + (16.00 g/mol x 8) = 811.9 g/mol

Finally, we can calculate the mass of the precipitate formed:
mass = moles x molar mass
mass = 0.0015 moles x 811.9 g/mol = 1.2 g

Therefore, the mass of the precipitate that will form is 1.2 grams.

To find the mass of precipitate formed in the reaction between sodium phosphate and lead(II) nitrate, we need to determine the limiting reactant. The limiting reactant is the one that is completely consumed, thus determining the maximum amount of product that can be formed.

First, let's write the balanced equation for the reaction between sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2):

3 Na3PO4 + 2 Pb(NO3)2 -> 6 NaNO3 + Pb3(PO4)2

From the balanced equation, we can see that the reaction ratio between sodium phosphate and lead(II) nitrate is 3:2.

Now, let's calculate the number of moles of sodium phosphate (Na3PO4) and lead(II) nitrate (Pb(NO3)2) in the given volumes:

Number of moles of sodium phosphate:
0.030 L * 0.30 mol/L = 0.009 mol

Number of moles of lead(II) nitrate:
0.020 L * 0.23 mol/L = 0.0046 mol

Next, let's determine the limiting reactant. To do that, we compare the reaction ratio of the two reactants:

Moles of sodium phosphate (Na3PO4) / 3 = Moles of lead(II) nitrate (Pb(NO3)2) / 2

0.009 mol / 3 = 0.0046 mol / 2

0.003 mol = 0.0023 mol

As we can see, the moles of lead(II) nitrate are lower than the moles of sodium phosphate, which means lead(II) nitrate is the limiting reactant.

Now, let's calculate the moles of lead(II) phosphate (Pb3(PO4)2) from the limiting reactant:

Moles of lead(II) phosphate (Pb3(PO4)2) = Moles of lead(II) nitrate (Pb(NO3)2)

0.0023 mol

Finally, let's calculate the molar mass of lead(II) phosphate and then use it to find the mass of the precipitate:

Molar mass of Pb3(PO4)2 = (207.2 g/mol) + (3 * (31.0 g/mol)) + (8 * (16.0 g/mol))
= 811.2 g/mol

Mass of precipitate = Moles of lead(II) phosphate (Pb3(PO4)2) * Molar mass of Pb3(PO4)2

Mass of precipitate = 0.0023 mol * 811.2 g/mol

Mass of precipitate = 1.866 g

Therefore, the mass of precipitate that will form is approximately 1.866 grams.

2Na3PO4 + 3Pb(NO3)2 ==> 6NaNO3 + Pb3(PO4)2

mols Na3PO4 = M x L
Using the coefficients in the balanced equation, convert mols Na3PO4 to mols Pb3(PO4)2.

Then convert mols ppt to grams. g = mols x molar mass.