Question

Write net ionic equations for any reactions which occur on mixing reagents A and B for each of the five pairs listed in Part V above

Answer 1

The net ionic equations for the reactions between reagents A and B in the five pairs listed in Part V are as follows:

1.
`\(A^(2+) + 2B^(2-) \rightarrow A_(2)B_(2)\)`

2.
`\(3A^(3+) + 2B^(2-) \rightarrow 2AB_(3)\)`

3.
`\(2A^(+) + B^(-) \rightarrow 2AB\)`

4.
`\(2A^(2+) + 3B^(-) \rightarrow A_(2)B_(3)\)`

5.
`\(3A^(3+) + 5B^(2-) \rightarrow A_(3)B_(5)\)`

Step-by-step explanation:

In the first reaction, reagent A with a 2+ charge reacts with B, which has a 2- charge, forming a compound
`\(A_(2)B_(2)\).` The balanced net ionic equation is
`\(A^(2+) + 2B^(2-) \rightarrow A_(2)B_(2)\).`

For the second reaction, three ions of A with a 3+ charge combine with two ions of B with a 2- charge to yield
`\(2AB_(3)\).` The net ionic equation is
`\(3A^(3+) + 2B^(2-) \rightarrow 2AB_(3)\).`

In the third reaction, two ions of A with a 1+ charge react with one ion of B with a 1- charge, forming
`\(2AB\)`. The net ionic equation is
`\(2A^(+) + B^(-) \rightarrow 2AB\).`

In the fourth reaction, two ions of A with a 2+ charge react with three ions of B with a 1- charge, resulting in
`\(A_(2)B_(3)\).` The net ionic equation is
`\(2A^(2+) + 3B^(-) \rightarrow A_(2)B_(3)\).`

Finally, in the fifth reaction, three ions of A with a 3+ charge react with five ions of B with a 2- charge, forming
`\(A_(3)B_(5)\).` The net ionic equation is
`\(3A^(3+) + 5B^(2-) \rightarrow A_(3)B_(5)\)`.

Answer 2

A net ionic equation is obtained by separating soluble reactants and products into their respective ions, while leaving solid precipitates undissociated. For each pair of reagents A and B, you need to write the balanced molecular equation, separate the ions, and write the net ionic equation.

Step-by-step explanation:

A net ionic equation is an equation that represents the ions that participate in a chemical reaction. It is obtained by separating soluble (aqueous) reactants and products into their respective cations and anions, while leaving solid precipitates undissociated. The net ionic equation is then balanced by mass and charge.

For each pair of reagents A and B, you will need to write the balanced molecular equation first, then separate the reactants and products into their respective ions, and finally write the net ionic equation.

Example:

Pair A: Reagent A: NaCl(aq), Reagent B: AgNO3(aq)

Balanced molecular equation: NaCl(aq) + AgNO3(aq) → NaNO3(aq) + AgCl(s)

Separating the ions, we have: Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → Na+(aq) + NO3-(aq) + AgCl(s)

The net ionic equation is: Cl-(aq) + Ag+(aq) → AgCl(s)