Chemical Reactions And Equation Balance

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Identify and classify the chemical reactions provided, balance the equations correctly, and determine the type of each reaction (synthesis, decomposition, single displacement, double displacement, combustion, etc.).

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Chemistry is fundamentally based on understanding chemical reactions and equations, which describe how substances interact and transform into new compounds. Accurate representation and classification of these reactions are crucial for mastering chemical concepts and applying them in real-world contexts. The provided reactions encompass various types of chemical interactions, each demonstrating different mechanisms of chemical change.

1. Synthesis Reaction: Co(s) + O₂(g) → Co₂O₃(s)

This reaction showcases a synthesis process where cobalt metal reacts with oxygen to form cobalt(III) oxide. To balance the reaction, ensure the atoms of each element are equal on both sides.

Unbalanced: Co(s) + O₂(g) → Co₂O₃(s)

Balance cobalt (Co): 2 Co atoms on the right, so place a coefficient of 2 in front of Co on the left: 2 Co + O₂ → Co₂O₃

Balance oxygen: 3 O atoms in Co₂O₃, but O₂ molecules contain 2 oxygen atoms each. To balance oxygen, multiply O₂ by 3: 2 Co + 3 O₂ → 2 Co₂O₃, which results in 6 oxygen atoms on the reactant side, but the product has 3×2=6 oxygen atoms, so multiply Co₂O₃ by 2: 2 Co + 3 O₂ → 2 Co₂O₃

Final balanced equation: 4 Co + 3 O₂ → 2 Co₂O₃

This is a synthesis reaction, as two reactants form a single product.

2. Decomposition Reaction: LiClO₃(s) → LiCl(s) + O₂(g)

This reaction involves the breakdown of lithium chlorate into lithium chloride and oxygen gas. It’s a typical decomposition process, often facilitated by heating.

Balance lithium (Li): LiClO₃ → LiCl — lithium already balanced.

Balance chlorine (Cl): Cl is balanced with 1 on each side.

Balance oxygen (O): 3 oxygens in lithium chlorate break into O₂ molecules. 3 oxygen atoms produce 1.5 O₂ molecules, but fractional coefficients are avoided by multiplying through by 2:

2 LiClO₃ → 2 LiCl + 3 O₂

This reaction is a decomposition, where a single compound breaks into two products.

3. Single Displacement Reaction: Cu(s) + AgC₂H₃O₂(aq) → Cu(C₂H₃O₂)₂(aq) + Ag(s)

This reaction depicts copper displacing silver from silver acetate, characteristic of single displacement reactions, which involve one element replacing another in a compound.

Balance copper (Cu): exists as a single atom on both sides.

Balance silver (Ag): only one on each side, both balanced.

Balance acetate groups: Cu forms Cu(C₂H₃O₂)₂, indicating two acetate ions coordinate with copper; thus, 2 AgC₂H₃O₂ react with 1 Cu to produce Cu(C₂H₃O₂)₂ and Ag:

Cu(s) + 2 AgC₂H₃O₂(aq) → Cu(C₂H₃O₂)₂(aq) + 2 Ag(s)

Classified as a single displacement reaction, with copper replacing silver.

4. Double Displacement Reaction: Pb(NO₃)₂(aq) + LiCl(aq) → PbCl₂(s) + LiNO₃(aq)

This involves the exchange of ions between two aqueous solutions leading to the formation of a precipitate, lead chloride, which confirms a double displacement or double replacement reaction.

Balance lead (Pb): 1 on both sides.

Balance nitrate (NO₃): 2 on the reactant side, 2 on the product side, balanced.

Balance lithium (Li) and chloride (Cl): 1 each on reactant sides; 1 each on products.

Balanced equation: Pb(NO₃)₂(aq) + 2 LiCl(aq) → PbCl₂(s) + 2 LiNO₃(aq)

5. Acid-Base Reaction: H₂SO₄(aq) + Al(OH)₃(aq) → Al₂(SO₄)₃(aq) + H₂O(l)

In this acid-base neutralization, sulfuric acid reacts with aluminum hydroxide to produce aluminum sulfate and water.

Balance aluminum (Al): 2 in aluminum sulfate, so place a coefficient of 2 in front of Al(OH)₃:

H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + H₂O

Balance sulfate (SO₄): 3 in aluminum sulfate, so 3 H₂SO₄ molecules are required:

3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O

This is a typical acid-base neutralization reaction.

6. Precipitation Reaction: HCIO₄(aq) + Ba(OH)₂(s) → Ba(ClO₄)₂(s) + H₂O(l)

Here, barium hydroxide reacts with perchloric acid to form barium perchlorate precipitate and water.

Balance barium (Ba): 1 on both sides; chlorine (Cl): 2 in Ba(ClO₄)₂; thus, 2 HCIO₄ molecules are involved:

2 HCIO₄ + Ba(OH)₂ → Ba(ClO₄)₂ + 2 H₂O

This is a precipitation reaction involving ionic compounds.

7. Redox Reaction: Co(NO₃)₂(aq) + H₂S(g) → CoS(s) + HNO₃(aq)

This reaction involves oxidation-reduction processes, where cobalt(II) nitrate reacts with hydrogen sulfide to produce cobalt sulfide precipitate and nitric acid.

The cobalt is reduced, and sulfur in H₂S is oxidized to sulfide, characteristic of redox reactions.

Unbalanced: Co(NO₃)₂ + H₂S → CoS + HNO₃

Balance nitrogen and nitrate: 2 nitrates on the reactant side, so:

Co(NO₃)₂ + H₂S → CoS + 2 HNO₃

8. Single Displacement or Redox: Fe(s) + Cd(NO₃)₂(aq) → Fe(NO₃)₃(aq) + Cd(s)

Iron displaces cadmium in solution, producing iron(III) nitrate and solid cadmium, depicting a redox process with electron transfer.

Balance atoms: Fe and Cd are balanced; nitrates: 3 NO₃ groups in Fe(NO₃)₃, so 3 Cd nitrates are involved:

Fe(s) + 3 Cd(NO₃)₂(aq) → Fe(NO₃)₃(aq) + 3 Cd(s)

9. Decomposition: Fe₂(CO₃)₃(s) → Fe₂O₃(s) + CO₂(g)

This is a decomposition reaction where iron(III) carbonate breaks down upon heating into iron(III) oxide and carbon dioxide gas.

Unbalanced: Fe₂(CO₃)₃ → Fe₂O₃ + CO₂

Balanced: The original formula is already balanced.

10. Synthesis Reaction: Sn(s) + P(s) → Sn₃P₂(s)

This reaction involves the combination of tin and phosphorus to form tin phosphide, a synthesis process.

Balance tin (Sn): 3 on the product side, so 3 Sn atoms react:

Sn + P → Sn₃P₂

Balance phosphorus (P): 2 atoms on the product side, so multiply P by 2/2 or use fractional coefficients, then scale to whole numbers: 3 Sn + 2 P → Sn₃P₂

This is a synthesis reaction where multiple elements combine to form a compound.

Conclusion

Each reaction has been carefully classified and balanced, illustrating the fundamental types of chemical reactions including synthesis, decomposition, single and double displacement, redox, acid-base, and precipitation reactions. Understanding these reactions enhances our ability to predict product formation, energy changes, and reaction conditions in practical chemistry applications. Proper balancing ensures conservation of mass, essential for the accurate stoichiometric calculations necessary in laboratory and industrial processes.

References

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