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18 February, 13:29

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.

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  1. 18 February, 14:58
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    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
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