Question #69d7d

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Question #69d7d

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Answer 1 · 2017-03-23T01:34:54

$3 {Mg}_{(s)} + N_{2 (g)} \to {Mg}_{3} N_{2 (s)}$ $3"Mg"_ ((s)) + "N"_ (2(g)) -> "Mg"_ 3"N"_ (2(s))$

Explanation:

Magnesium will react with nitrogen gas to produce magnesium nitride, ${Mg}_{3} N_{2}$ $"Mg"_3"N"_2$ , and ionic compound that consists of magnesium cations, ${Mg}^{2 +}$ $"Mg"^(2+)$ , and nitride anions, $N^{3 -}$ $"N"^(3-)$ .

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The unbalanced chemical equation looks like this

${Mg}_{(s)} + N_{2 (g)} \to {Mg}_{3} N_{2 (s)}$ $"Mg"_ ((s)) + "N"_ (2(g)) -> "Mg"_ 3"N"_ (2(s))$

Notice that all you need to do in order to balance this chemical equation is multiply magnesium metal by $3$ $3$ . This will give you a total of $3$ $3$ atoms of magnesium and $2$ $2$ atoms of nitrogen on both sides of the equation.

$3 {Mg}_{(s)} + N_{2 (g)} \to {Mg}_{3} N_{2 (s)}$ $3"Mg"_ ((s)) + "N"_ (2(g)) -> "Mg"_ 3"N"_ (2(s))$

This reaction is an example of a redox reaction. Here magnesium is being oxidized and nitrogen gas is being reduced.

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Question #69d7d

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