Question

When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)2 produced when 1.2 mL of 0.152 M Ba(NO3)2 and 4.2 mL of 0.604 M K3PO4 are mixed. Calculate the mass to 3 significant figures.

Answer 1

Answer:

Mass of Ba₃(PO₄)₂ = 0.0361 g

Explanation:

Given data:

Volume of Ba(NO₃)₂ = 1.2 mL (1.2 × 10⁻³ L )

Molarity of Ba(NO₃)₂ = 0.152 M

Volume of K₃PO₄ = 4.2 mL (4.2 × 10⁻³ L)

Molarity of K₃PO₄ =  0.604 M

Mass of Ba₃(PO₄)₂ produced = ?

Solution:

Chemical equation:

3Ba(NO₃)₂  + 2K₃PO₄  → Ba₃(PO₄)₂  + 6KNO₃

Number of moles of Ba(NO₃)₂ = Molarity × Volume in litter

Number of moles of Ba(NO₃)₂ = 0.152 M × 1.2 × 10⁻³ L

Number of moles of Ba(NO₃)₂ = 0.182 × 10⁻³ mol

Number of moles of K₃PO₄ = Molarity × Volume in litter

Number of moles of K₃PO₄ = 0.604 M × 4.2 × 10⁻³ L

Number of moles of K₃PO₄ = 2.537 × 10⁻³ mol

Now we will compare the moles of Ba₃(PO₄)₂ with K₃PO₄ and Ba(NO₃)₂ .

              Ba(NO₃)₂        :         Ba₃(PO₄)₂

                   3                :               1

              0.182 × 10⁻³    :              1/3 ×0.182 × 10⁻³ = 0.060 × 10⁻³ mol

                K₃PO₄           :          Ba₃(PO₄)₂

                   2                 :                1

              2.537 × 10⁻³     :               1/2 ×  2.537 × 10⁻³= 1.269 × 10⁻³ mol

The number of moles of Ba₃(PO₄)₂ produced by  Ba(NO₃)₂  are less it will limiting reactant.

Mass of Ba₃(PO₄)₂ = moles × molar mass

Mass of Ba₃(PO₄)₂ = 0.060 × 10⁻³ mol × 601.93 g/mol

Mass of Ba₃(PO₄)₂ = 36.12 × 10⁻³ g

Mass of Ba₃(PO₄)₂ = 0.0361 g

Answer 2

The mass of Ba₃(PO₄)₂ produced when 1.2 mL of 0.152 M Ba(NO3)2 and 4.2 mL of 0.604 M K3PO4 are mixed is 0.0366 g

First, we will write the balanced chemical equation for the reaction

2K₃PO₄ + 3Ba(NO₃)₂ → Ba₃(PO₄)₂ + 6KNO₃

This means 2 moles of K₃PO₄ will react with 3 moles of Ba(NO₃)₂ to produce 1 mole of Ba₃(PO₄)₂ and 6 moles KNO₃

To calculate the mass, in grams, of the Ba₃(PO₄)₂ produced,

First, we will determine the number of moles of each reactant present

• For K₃PO₄

Volume = 4.2 mL = 0.0042 L

Concentration = 0.604 M

From the formula

Number of moles = Concentration × Volume

Number of moles of K₃PO₄ = 0.604 × 0.0042

Number of moles of K₃PO₄ = 0.0025368 moles

• For Ba(NO₃)₂

Volume = 1.2mL = 0.0012 L

Concentration = 0.152 M

∴ Number of moles of Ba(NO₃)₂ = 0.152 × 0.0012

Number of moles of Ba(NO₃)₂ = 0.0001824 moles

Since 2 moles of K₃PO₄ will react with 3 moles of Ba(NO₃)₂

Then, $\frac{2}{3}\times 0.0001824$ moles of K₃PO₄ will react with 0.001824 moles of Ba(NO₃)₂

$\frac{2}{3}\times 0.0001824= 0.0001216$

∴ Only 0.0001216 moles of K₃PO₄ reacted

(NOTE: K₃PO₄ is the excess reagent while Ba(NO₃)₂ is the limiting reagent)

From the equation of reaction,

2 moles of K₃PO₄ will react with 3 moles of Ba(NO₃)₂ to produce 1 mole of Ba₃(PO₄)₂

Then,

0.0001216 moles of K₃PO₄ will react with 0.001824 moles of Ba(NO₃)₂ to produce $\frac{0.0001216}{2}$ moles of Ba₃(PO₄)₂

$\frac{0.0001216}{2} = 0.0000608$

∴ 0.0000608 moles of Ba₃(PO₄)₂ was produced

Now, we will determine the mass of Ba₃(PO₄)₂ produced

Using the formula

Mass = Number of moles × Molar mass

Number of moles of Ba₃(PO₄)₂ = 0.0000608 moles

Molar mass of Ba₃(PO₄)₂ = 601.93 g/mol

∴ Mass of Ba₃(PO₄)₂ produced = 0.0000608 × 601.93

Mass of Ba₃(PO₄)₂ produced = 0.036597 g

Mass of Ba₃(PO₄)₂ produced ≅ 0.0366 g

Hence, the mass of Ba₃(PO₄)₂ produced when 1.2 mL of 0.152 M Ba(NO3)2 and 4.2 mL of 0.604 M K3PO4 are mixed is 0.0366 g

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