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CaHeK987 [17]
3 years ago
5

What will create a insoluble salt with a potassium ion

Chemistry
1 answer:
navik [9.2K]3 years ago
4 0

Explanation:

1) The dissolution of the salt potassium sulfite:

K₂SO₃(aq) → 2K⁺(aq) + SO₃²⁻(aq).

Potassium has +1 charge because it lost one electron to accomplish stabile electron configuration of noble gas argon.

2) From dissolution reaction: n(K⁺) : n(SO₃²⁻) = 2 : 1.

n(K⁺) = 0.700 mol.

0.700 mol : n(SO₃²⁻) = 2 : 1.

n(SO₃²⁻) = 0.700 mol ÷ 2.

n(SO₃²⁻) = 0.350 mol; amount of sulfite anions.

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Complete and balance the following reaction by filling in the missing coefficients. Assume the reaction is occurring in a basic,
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Answer:

4 MnO₄⁻(aq) + 3 CH₃CH₂OH(aq) ⟶ 4 MnO₂(s) + 1 OH⁻(aq) + 3 CH₃COO⁻(aq) + 4 H₂O(l)

Explanation:

To balance a redox reaction we use the ion-electron method.

Step 1: Identify both half-reactions

Reduction: MnO₄⁻(aq) ⟶ MnO₂(s)

Oxidation: CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq)

Step 2: Balance the mass adding H₂O and OH⁻ where necessary.

2 H₂O(l) + MnO₄⁻(aq) ⟶ MnO₂(s) + 4 OH⁻(aq)

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Step 3: Balance the charge adding eelctrons where necessary.

2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq)

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Step 4: Multiply both half-reactions by numbers that assure that the number of electrons gained and lost are the same.

4 × (2 H₂O(l) + MnO₄⁻(aq) + 3 e⁻ ⟶ MnO₂(s) + 4 OH⁻(aq))

3 × (5 OH⁻(aq) + CH₃CH₂OH(aq) ⟶ CH₃COO⁻(aq) + 4 H₂O(l) + 4 e⁻)

Step 5: Add both half-reactions and cancel what is repeated.

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