Question #9701f

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Question #9701f

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Answer 1 · 2016-05-05T00:58:02

$C_{(s)} + 4 {HNO}_{3 (a q)} \to 2 H_{2} O_{(l)} + 4 {NO}_{2 (a q)} + {CO}_{2 (g)} ↑ ⏐$ $"C"_ ((s)) + 4"HNO"_ (3(aq)) -> 2"H"_ 2"O"_ ((l)) + 4"NO"_ (2(aq)) + "CO"_(2(g)) uarr$

Explanation:

You're dealing with a redox reaction in which carbon reduces hot, concentrated nitric acid to nitrogen dioxide, ${NO}_{2}$ $"NO"_2$ , while being oxidized to carbon dioxide, ${CO}_{2}$ $"CO"_2$ , in the process.

The balanced chemical equation that describes this reaction looks like this

${0 C}_{(s)} + 4 H + 5 N O_{3 (a q)} \to 2 H_{2} O_{(l)} + 4 + 4 N O_{2 (a q)} + + 4 C O_{2 (g)}$ $stackrel(color(blue)(0))("C")_ ((s)) + 4"H"stackrel(color(blue)(+5))("N")"O"_ (3(aq)) -> 2"H"_ 2"O"_ ((l)) + 4stackrel(color(blue)(+4))("N")"O"_ (2(aq)) + stackrel(color(blue)(+4))("C")"O"_(2(g))$ $↑$ $uarr$

Here carbon is being oxidized from a $0$ $color(blue)(0)$ oxidation state in elemental carbon to a $+ 4$ $color(blue)(+4)$ oxidation state in carbon dioxide.

Nitrogen, on the other hand, is being reduced from a $+ 5$ $color(blue)(+5)$ oxidation state in nitric acid to a $+ 4$ $color(blue)(+4)$ oxidation state in nitrogen dioxide.

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Question #9701f

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