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shepuryov [24]
3 years ago
9

In addition to mass balance, oxidation-reduction reactions must be balanced such that the number of electrons lost in the oxidat

ion equals the number of electrons gained in the reduction. This balancing can be done by two methods: the half-reaction method or the oxidation number method. The half-reaction method balances the electrons lost in the oxidation half-reaction with the electrons gained in the reduction half-reaction. In either method H2O(l), OH−(aq), and H+(aq) may be added to complete the mass balance. Which substances are used depends on the reaction conditions.
Chemistry
1 answer:
Vaselesa [24]3 years ago
4 0

Answer:

OH−(aq), and H+(aq)

Explanation:

Redox reactions may occur in acidic or basic environments. Usually, if a reaction occurs in an acidic environment, hydrogen ions are shown as being part of the reaction system. For instance, in the reduction of the permanganate ion;

MnO4^-(aq) + 8H^+(aq) +5e-------> Mn^2+(aq) + 4H2O(l)

The appearance of hydrogen ion in the reaction equation implies that the process takes place under acidic reaction conditions.

For reactions that take place under basic conditions, the hydroxide ion is part of the reaction equation.

Hence hydrogen ion and hydroxide ion are included in redox reaction half equations depending on the conditions of the reaction whether acidic or basic.

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Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

nitrogen     31.8%         =     31.8 g

Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

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