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Unlike other Group 2A(2) metals, beryllium reacts like aluminum and zinc with concentrated aqueous base to release hydrogen gas and form oxoanions of formula M(OH) _4 ⁿ⁻. Write equations for the reactions of these three metals with NaOH.

Respuesta :

2Al + 2H2O + 2NaOH = 2Al + 2H2O + 2NaAlO2 (reaction with aluminum)

Zn+2NaOH→Na2ZnO2+H2. (Reaction with zinc)

2NaOH + Be + Na2BeO2 + H2 (reaction with beryllium)

Sodium aluminate (NaAlO2) and hydrogen gas are the byproducts of the reaction between aluminum and aqueous sodium hydroxide (NaOH). Because it is an amphoteric element, aluminum reacts with bases.

2Al + 2H2O + 2NaOH = 2Al + 2H2O + 2NaAlO2

Excess sodium hydroxide and zinc combine to generate sodium zincate (Na2ZnO2) and hydrogen gas. With the release of hydrogen gas, zinc interacts with NaOH to produce sodium zincate.

Zn+2NaOH→Na2ZnO2+H2.

Sodium beryllate and hydrogen are produced when beryllate and sodium hydroxide combine. The temperature where this reaction occurs is between 400 and 500 °C.

2NaOH + Be + Na2BeO2 + H2

Learn more about amphoteric here-

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