Make up the equations for the reactions of interaction: a) aluminum with sulfur; b) iron with hydrochloric acid solution; c) sodium with bromine. In the reaction equation (a), arrange the coefficients using the electronic balance method.
A) 2Al + 3S = Al2 S3.
1. We arrange the oxidation states for each element.
Al 0+ S 0 = Al +32 S-23
2. Write out the elements that have changed their oxidation state.
We determine how many electrons they gave or received (we subtract the final one from the initial oxidation state).
The aluminum from the zero degree turned into an aluminum ion with the degree + 3. He gave up 3 electrons, that is, he is a reducing agent, he himself is oxidized.
Sulfur from the zero degree has turned into an ion with a degree of -2, that is, it has taken 2 electrons, is an oxidizing agent, while it itself is reduced.
B) Fe + 2HCl = FeCl2 + H2 (iron (II) chloride is formed)
С) 2Na + Br2 = 2NaBr (sodium bromide).
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