Question

Based on the replacement reaction, what would the products of the reaction be?

BeSO4 + 2NH4OH → Be()2 + (NH4)2SO4

Answer 1

OH

Step-by-step explanation:

Answer 2

`\rm Be(OH)_2` and
`\rm (NH_4)_2 SO_4`. The missing ion would be
`\rm OH^(-)`.

Step-by-step 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:

• 
  `\rm BeSO_4`, and
• 
  `\rm NH_4 OH`.

In particular,

• 
  `\rm BeSO_4` is made up of
  `\rm Be^(2+)` ions and
  `\rm {SO_4}^(2-)` ions, while
• 
  `\rm NH_4 OH` is made up of
  `\rm {NH_4}^(+)` ions and
  `\rm OH^(-)` ions.

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

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

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

Hence, the two products will be
`\rm Be(OH)_2` and
`\rm (NH_4)_2 SO_4`.

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