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\text{Al} +Al+start text, A, l, end text, plus \text{HCl} \rightarrowHCl→start text, H, C, l, end text, right arrow \text{AlCl}_3 +AlCl 3 ​ +start text, A, l, C, l, end text, start subscript, 3, end subscript, plus \text{H}_2H 2 ​ start text, H, end text, start subscript, 2, end subscript Stuck?Watch a video or use a hint.

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Answer:

2Al(s) + 6HCl(l)  ⇒ 2AlCl3(s)  +    3H2(g)  

Explanation:

Al displaces H from HCl, because Al is higher than H in the electrochemical series, hence, more reactive than H.    

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