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Balance the equation below in acid solution. You need not include the state symbols.
Br⁻ + MnO₄⁻ → BrO₃⁻ + Mn²⁺



Answer :

zarahaider4211

Answer:

[tex]\text{4Br + 3MnO}_4 \rightarrow \text{4BrO}_3+3\text{Mn}[/tex]

Explanation:

Balancing Equations

Identifying the Charges of Transition Metals

Transition metals don't have a defined ionic charge, so the only way its charge from a compound is by looking at the other element's (or polyatomic) charge.

A compound, consisting of a cation (+ charged atom) and an anion ( - charged atom), must have a neutral charge.

So, the transition metal's charge has a value of the same magnitude as its non-transition metal counterpart with an opposite charge symbol.

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Determining the Coefficients to Balance an Equation

To balance an equation, the number of atoms of the same type of element (or polyatomic) must be the same on the reactant (left side) and product sides (right side).

The quickest method to solving these problems is by

  1. counting out the number of atoms of each element (look out for coefficients and subscripts)
  2. Multiply any part of the equation to balance out the metals
  3. repeat step 2 for non-metals (exclude hydrogen and oxygen)
  4. repeat step 2 for hydrogen and oxygen.

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Solving the Problem

First, we count up all the different atoms on both sides of the given equation

                         [tex]\text{Br}^- + \text{MnO}_4^- \rightarrow \text{BrO}_3^-+\text{Mn}^{2+}[/tex].

Reactant side:                      |                  Product side:

Br: 1                                       |         Br: 1

Mn: 1                                     |         Mn: 1

O: 4                                       |         O: 3

Everything is balanced except for oxygen, so we multiply each side by a factor that will make their number of atoms the same.

To determine that number we consider the greatest common factor (GCF) of 4 and 3: 12. Thus, we multiply the oxygen on the reactant side by 3 and the product side by 4.

                          [tex]\text{Br}^- + \text{3MnO}_4^- \rightarrow \text{4BrO}_3^-+\text{Mn}^{2+}[/tex]

This means the number of atoms in Mn and Br will also change.

Reactant side:                      |                  Product side:

Br: 1                                       |         Br: 4

Mn: 3                                     |         Mn: 1

O: 12                                      |         O: 12

Now, we must balance out Br and Mn.

Noticing that the Br on the reactant side is alone, we can multiply it by 4 to balance the Br element. A similar reasoning can be applied to multiplying the Mn on the product side by 3.

                          [tex]\text{4Br}^- + \text{3MnO}_4^- \rightarrow \text{4BrO}_3^-+\text{3Mn}^{2+}[/tex]

Reactant side:                      |                  Product side:

Br: 4                                       |         Br: 4

Mn: 3                                     |         Mn: 3

O: 12                                      |         O: 12

We can stop here as all the atoms are balanced.

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