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3 years ago

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Chemistry

1 answer:

lawyer [7]3 years ago

3 0

Answer:

A. 466 g of BaSO₄.

B. 2.25 moles of H₂O.

Explanation:

The balanced equation for the reaction is given below:

BaO + H₂SO₄ —> BaSO₄ + H₂O

Next, we shall determine the mass of H₂SO₄ that reacted and the mass of BaSO₄ produced from the balanced equation. This can be obtained as follow:

Molar mass of H₂SO₄ = (2×1) + 32 + (16×4)

= 2 + 32 + 64

= 98 g/mol

Mass of H₂SO₄ from the balanced equation = 1 × 98 = 98 g

Molar mass of BaSO₄ = 137 + 32 + (16×4)

= 137 + 32 + 64

= 233 g/mol

Mass of BaSO₄ from the balanced equation = 1 × 233 = 233 g

SUMMARY:

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 233 g of BaSO₄.

A. Determination of the mass of BaSO₄ produced by the reaction of 196 g of H₂SO₄.

From the balanced equation above,

98 g of H₂SO₄ reacted to produce 233 g of BaSO₄.

Therefore, 196 g of H₂SO₄ will react to produce = (196 × 233) /98 = 466 g of BaSO₄.

Thus, 466 g of BaSO₄ were obtained from the reaction.

B. Determination of the number of mole of H₂O produced by the reaction of 345 g of BaO.

We'll begin by calculating the number of mole in 345 g of BaO. This can be obtained as follow:

Mass of BaO = 345 g

Molar mass of BaO = 137 + 16 = 153 g/mol

Mole of BaO =?

Mole = mass /Molar mass

Mole of BaO = 345 / 153

Mole of BaO = 2.25 moles

Finally, we shall determine the number of mole of H₂O produced from the reaction. This can be obtained as follow:

BaO + H₂SO₄ —> BaSO₄ + H₂O

From the balanced equation above,

1 mole of BaO reacted to produce 1 mole of H₂O.

Therefore, 2.25 moles of BaO will also react to produce 2.25 moles of H₂O.

Thus, 2.25 moles of H₂O were obtained from the reaction.

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(c) [NH₃]

$c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}$

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 6.106×10⁻³ 0.1462 0

C/mol·L⁻¹: -6.106×10⁻³ 0.1462-6×6.106×10⁻³ 0

E/mol·L⁻¹: 0 0.1095 6.106×10⁻³

Then we approach equilibrium from the right.

Ni²⁺ + 6NH₃ ⇌ Ni(NH₃)₆²⁺

I/mol·L⁻¹: 0 0.1095 6.106×10⁻³

C/mol·L⁻¹: +x +6x -x

E/mol·L⁻¹: x 0.1095+6x 6.106×10⁻³-x

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}$

Kf is large, so x ≪ 6.106×10⁻³. Then

$K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}$

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