Question

When silver nitrate and sodium sulfate react, a precipitate forms. What mass of the precipitate is obtained when 100. ml of 0.100 M sodium sulfate is added to 150. ml of 0.100 M silver nitrate?

Answer 1

Answer:

2.338 g of precipitate are obtained

Explanation:

The reaction that takes place is:

• 2AgNO₃(aq) + Na₂SO₄ → Ag₂SO₄(s) + 2NaNO₃(aq)

To find out how much silver sulfate (Ag₂SO₄) is produced, we first determine the limiting reactant:

• 0.100 M Na₂SO₄ * 100 mL = 10 mmol Na₂SO₄

• 0.100 M AgNO₃ * 150 mL = 15 mmol AgNO₃

10 mmol of Na₂SO₄ would react completely with 20 mmol of AgNO₃. As there are not so many AgNO₃ moles, then AgNO₃ is the limiting reactant:

• 15 mmol AgNO₃ * 1 molAg₂SO₄/2 molAgNO₃ * 311.799mg/mmolAg₂SO₄ = 2338.49 mg Ag₂SO₄

• 2338.49 mg = 2.338 g Ag₂SO₄

Answer 2

The mass of the precipitate, Ag₂SO₄ obtained is 2.34 g

Determination of the mole of Na₂SO₄

• Volume of Na₂SO₄ = 100 mL = 100 / 1000 = 0.1 L

• Molarity of Na₂SO₄ = 0.1 M

• Mole of Na₂SO₄ =?

Mole = Molarity x Volume

Mole of Na₂SO₄ = 0.1 × 0.1

Mole of Na₂SO₄ = 0.01 mole

Determination of the mole of AgNO₃

• Volume of AgNO₃ = 150 mL = 150 / 1000 = 0.15 L

• Molarity of AgNO₃ = 0.1 M

• Mole of AgNO₃ =?

Mole = Molarity x Volume

Mole of AgNO₃ = 0.1 × 0.15

Mole of AgNO₃ = 0.015 mole

Determination of the limiting reactant

Na₂SO₄ + 2AgNO₃(aq) → Ag₂SO₄(s) + 2NaNO₃(aq)

From the balanced equation above,

1 mole of Na₂SO₄ reacted with 2 moles AgNO₃

Therefore,

0.01 mole of Na₂SO₄ will react with = 0.01 × 2 = 0.02 mole of AgNO₃

From the above calculation, we can see that a higher amount of AgNO₃ (i.e 0.02 mole) than what was given (i.e 0.015 mole) is needed to react completely with 0.01 mole of Na₂SO₄.

Therefore, AgNO₃ is the limiting reactant and Na₂SO₄ is the excess reactant.

Determination of the mole of the precipitate, Ag₂SO₄ obtained

Na₂SO₄ + 2AgNO₃(aq) → Ag₂SO₄(s) + 2NaNO₃(aq)

From the balanced equation above,

2 moles of AgNO₃ reacted to produce 1 mole of Ag₂SO₄.

Therefore,

0.015 mole of AgNO₃ will react to produce = 0.015 / 2 = 0.0075 mole of Ag₂SO₄

Determination of the mass of precipitate, Ag₂SO₄ obtained

• Mole of Ag₂SO₄ = 0.0075 mole

• Molar mass of Ag₂SO₄ = (2×108) + 32 + (16×4) = 312 g/mol

• Mass of Ag₂SO₄ =?

Mass = mole × molar mass

Mass of Ag₂SO₄ = 0.0075 × 312

Mass of Ag₂SO₄ = 2.34 g

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