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.

Mass of Ba₃ (PO₄) ₂ = 0.0361 g

Explanation:

Given dа ta:

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