An imaginary element (Atomic weight 93.7140) has three naturally-occurring isotopes with isotopic weights of 92.9469, 93.2923 an
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Answer:
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄+ 2H₂O
Explanation:
PbO₂ + MnSO₄ + HNO₃ ⟶ HMnO₄ + PbSO₄ + Pb(NO₃)₂ + H₂O
It will be easiest to balance this equation by the ion-electron method.
1. Write the ionic equation
PbO₂ + Mn²⁺ + SO₄²⁻ + H⁺ + NO₃⁻ ⟶ H⁺ + MnO₄⁻ + Pb²⁺ + SO₄²⁻ + Pb²⁺ + NO₃⁻ + H₂O
2. Eliminate H⁺, H₂O, and spectator ions
PbO₂ + Mn²⁺ ⟶ MnO₄⁻ + Pb²⁺
3. Separate the skeleton equation into two half-reactions.
PbO₂ ⟶ Pb²⁺
Mn²⁺ ⟶ MnO₄⁻
4. Balance all atoms other than H and O
Done
5. Balance O by adding water molecules to the deficient side
PbO₂ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻
6. Balance H by adding H⁺ ions to the deficient side.
PbO₂+ 4H⁺ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺
7. Balance charge by adding electrons to the deficient side.
PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O
Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e-
8. Multiply each half-reaction by a number to equalize the electrons transferred.
5 × [PbO₂+ 4H⁺ + 2e⁻ ⟶ Pb²⁺ + 2H₂O]
2 × [Mn²⁺ + 4H₂O ⟶ MnO₄⁻ + 8H⁺ + 5e⁻]
9. Add the two half-reactions.
5PbO₂+ 20H⁺ + 10e⁻ ⟶ 5Pb²⁺ + 10H₂O
<u> 2Mn²⁺ + 8H₂O ⟶ 2MnO₄⁻ + 16H⁺ + 10e⁻ </u>
5PbO₂ + 2Mn² + 8H₂O + 20H⁺ + 10e⁻⟶ 5Pb²⁺ + 2MnO₄⁻ + 10H₂O + 16H⁺ + 10e⁻
10. Cancel species that occur on each side of the equation
5PbO₂ + 2Mn² + <u>8H₂O</u> + <u>20H⁺</u> + <u>10e⁻</u> ⟶ 5Pb²⁺ + 2MnO₄⁻ + <u>10H₂O</u> + <u>16H⁺</u> + <u>10e⁻ </u>
becomes
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
11. Add the missing spectator ions
5PbO₂ + 2Mn²⁺ + 4H⁺ ⟶ 5Pb²⁺ + 2MnO₄⁻ + 2H₂O
+ 2SO₄²⁻ + 4NO₃⁻ + 2H⁺ + 2NO₃⁻ +2SO₄²⁻ + 6NO₃⁻ + 2H⁺
becomes
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
12. Check that all atoms are balanced.
Everything checks. The balanced equation is
5PbO₂ + 2MnSO₄ + 6HNO₃ ⟶ 2PbSO₄ + 3Pb(NO₃)₂ + 2HMnO₄ + 2H₂O
