<h3>
Answer:</h3>
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
<h3>
Explanation:</h3>
Concept tested: Balancing of chemical equations
- A chemical equation is balanced by putting appropriate coefficients on the products and reactants of the equation.
- Balancing chemical equations ensures that chemical equations obey law of conservation of mass.
- In this case; to balance the above equation we put the coefficients, 1, 3, 2, and 3 on the reactants and products.
- Therefore; the balanced chemical equation for the reaction is;
Fe₂O₃(s) + 3CO(g) → 2Fe(s) + 3CO₂(g)
Number of moles = mass of Ni /molecular mass of Ni
mass of nickel = 86.4 g
molecular mass of nickel = 58.69
number of moles of Ni in 86.4 g
=86.4/58.69
=1.472 mol
(rounded to four significant figures instead of three because the first digit of the answer starts with a 1).
Answer: shining a bright light on the object and testing for decomposition
Decomposition is the separation of a molecule into its elements or simpler compounds.
And a chemical property is the ability of a
substance to combine or to become one or more substances.
Then, testing the ability of an object for decomposition is testing a chemical property.
What is the percent composition by mass of oxygen in magnesium oxide, MgO?
Answer: 39.7 percent
Answer:
2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺
Explanation:
To balance a redox reaction in an acidic medium, we simply follow some rules:
- Split the reaction into an oxidation and reduction half.
- By inspecting, balance the half equations with respect to the charges and atoms.
- In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
- Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
- Multiply both equations with appropriate factors to balance the electrons in the two half equations.
- Add up the balanced half equations and cancel out any specie that occur on both sides.
- Check to see if the charge and atoms are balanced.
Solution
Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺
The half equations:
Zn → Zn²⁺ Oxidation half
MnO₄⁻ → Mn²⁺ Reduction half
Balancing of atoms(in acidic medium)
Zn → Zn²⁺
MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O
Balancing of charge
Zn → Zn²⁺ + 2e⁻
MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O
Balancing of electrons
Multiply the oxidation half by 5 and reduction half by 2:
5Zn → 5Zn²⁺ + 10e⁻
2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O
Adding up the two equations gives:
5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O
The net equation gives:
5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O
