How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas.

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How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas.

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Answer 1 · 2014-05-13T17:15:41

$2 Al (s) + 6 HCl (a q) \to 3 H_{2} (g) + 2 {AlCl}_{3} (a q)$ $color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))$

Explanation:

This reaction is between a metal and an acid which typically results in a salt and the release of hydrogen gas. The unbalanced reaction is

$A l + H C l \to H_{2} + A l C l_{3}$ $Al + HCl -> H_2 + AlCl_3$ .

This is a redox reaction, whose half-reactions are and become:

$2 (Al (s) \to {Al}^{3 +} (a q) + 3 e^{-})$ $2("Al"(s) -> "Al"^(3+)(aq) + cancel(3e^(-)))$
$3 (2 H^{+} (a q) + 2 e^{-} \to H_{2} (g))$ $3(2"H"^(+)(aq) + cancel(2e^(-)) -> "H"_2(g))$
$-----------------------------------------------$ $"-----------------------------------------------"$
$2 Al (s) + 6 H^{+} (a q) \to 3 H_{2} (g) + 2 {Al}^{3 +} (a q)$ $2"Al"(s) + 6"H"^(+)(aq) -> 3"H"_2(g) + 2"Al"^(3+)(aq)$

Aluminum oxidizes as $Al \to {Al}^{3 +}$ $"Al" -> "Al"^(3+)$ , while hydrogen reduces as $2 H^{+} \to H_{2}^{0}$ $2"H"^(+) -> "H"_2^0$ .

If we add back the spectator ${Cl}^{-}$ $"Cl"^(-)$ , we get:

$2 Al (s) + 6 HCl (a q) \to 3 H_{2} (g) + 2 {AlCl}_{3} (a q)$ $color(blue)(2"Al"(s) + 6"HCl"(aq) -> 3"H"_2(g) + 2"AlCl"_3(aq))$

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How can I balance this chemical equations? Aluminum and hydrochloric acid react to form aluminum chloride and hydrogen gas.

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