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3 years ago

27/13Al+4/2He >>>> 30/15P+

Chemistry

1 answer:

Veseljchak [2.6K]3 years ago

5 0

Answer

Explanation:

27/13Al+4/2He >>>> 30/15P+

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When aqueous solutions of K3PO4 and Ba(NO3)2 are combined, Ba3(PO4)2 precipitates. Calculate the mass, in grams, of the Ba3(PO4)

Firdavs [7]

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

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