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Based on the replacement reaction, what would the products of the reaction be?
BeSO4 + 2NH4OH → Be()2 + (NH4)2SO4

Respuesta :

jacob193

Answer:

[tex]\rm Be(OH)_2[/tex] and [tex]\rm (NH_4)_2 SO_4[/tex]. The missing ion would be [tex]\rm OH^{-}[/tex].

Explanation:

In a double replacement reaction, two ionic compounds exchange their ions to produce two different ionic compounds.

In this question, the two ionic compounds are:

  • [tex]\rm BeSO_4[/tex], and
  • [tex]\rm NH_4 OH[/tex].

In particular,

  • [tex]\rm BeSO_4[/tex] is made up of [tex]\rm Be^{2+}[/tex] ions and [tex]\rm {SO_4}^{2-}[/tex] ions, while
  • [tex]\rm NH_4 OH[/tex] is made up of [tex]\rm {NH_4}^{+}[/tex] ions and [tex]\rm OH^{-}[/tex] ions.

In a binary ionic compound, cations (positive ions) can only bond to anions (negative ions.)

  • [tex]\rm Be^{2+}[/tex] is a cation. In [tex]\rm BeSO_4[/tex], [tex]\rm Be^{2+}[/tex] was bounded [tex]\rm {SO_4}^{2-}[/tex] anions. During the reaction, it bonds with [tex]\rm OH^{-}[/tex] anions to produce [tex]\rm Be(OH)_2[/tex].
  • [tex]\rm {NH_4}^{+}[/tex] is also a cation. In [tex]\rm NH_4 OH[/tex], [tex]\rm {NH_4}^{+}[/tex] was bounded to [tex]\rm OH^{-}[/tex] ions. During the reaction, it bonds with [tex]\rm {SO_4}^{2-}[/tex] anions to produce [tex]\rm (NH_4)_2 SO_4[/tex].

Hence, the two products will be [tex]\rm Be(OH)_2[/tex] and [tex]\rm (NH_4)_2 SO_4[/tex].

Note that charges on the ions must balance. For example, a [tex]\rm Be^{2+}[/tex] ion carries twice as much charge as an [tex]\rm {NH_4}^{+}[/tex] ion. As a result, each [tex]\rm Be^{2+}[/tex] ion would bond with twice as many [tex]\rm OH^{-}[/tex] ions as [tex]\rm {NH_4}^{+}[/tex] would in [tex]\rm NH_4 OH[/tex].

jordonmcgavin

Answer:

OH

Explanation:

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