Question

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

Answer 1

Answer:

OH Would be the answer.

Explanation:

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Answer 2

Answer:

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

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 $\rm NH_4 OH$.