Question

50.0 mL of .100 M Na3PO4 is mixed with 150.0mL of .250 M Pb(NO3)2 to produce a solid precipitate of lead (II) Phosphate. What mass of this precipitate will be produced?

Answer 1

The mass of lead (II) phosphate produced in the reaction is 2.03 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$     .....(1)

• For sodium phosphate:

Molarity of sodium phosphate solution = 0.100 M

Volume of solution = 50.0 mL

Putting values in equation 1, we get:

$0.100M=\frac{\text{Moles of sodium phosphate}\times 1000}{50}\\\\\text{Moles of sodium phosphate}=\frac{0.100\times 50}{1000}=0.005mol$

• For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 0.250 M

Volume of solution = 150.0 mL

Putting values in equation 1, we get:

$0.250M=\frac{\text{Moles of lead (II) nitrate}\times 1000}{150}\\\\\text{Moles of lead (II) nitrate}=\frac{0.250\times 150}{1000}=0.0375mol$

The chemical equation for the reaction of sodium phosphate and lead (II) nitrate follows:

$3Pb(NO_3)_2+2Na_3PO_4\rightarrow Pb_3(PO_4)_2+6NaNO_3$

By Stoichiometry of the reaction:

2 moles of sodium phosphate reacts with 3 moles of lead (II) nitrate

So, 0.005 moles of sodium phosphate will react with = $\frac{3}{2}\times 0.005=0.0075mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, sodium phosphate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of sodium phosphate produces 1 mole of lead (II) phosphate

So, 0.005 moles of sodium phosphate will produce = $\frac{1}{2}\times 0.005=0.0025moles$ of lead (II) phosphate

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Molar mass of lead (II) phosphate = 811.54 g/mol

Moles of lead (II) phosphate = 0.0025 moles

Putting values in above equation, we get:

$0.0025mol=\frac{\text{Mass of lead (II) phosphate}}{811.54g/mol}\\\\\text{Mass of lead (II) phosphate}=(0.0025mol\times 811.54g/mol)=2.03g$

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