Iron metal is dissolved in dilute aqueous HCl. The resulting solution is warmed with dilute nitric acid and then treated with aqueous potassium iodide. Write equations for the reactions.

Ok so for the first step I got:
Fe + 2HCl --> FeCl2 + H2

I'm guessing that when nitric acid is added the Fe2+ will be oxidised to Fe3+, but I don't know whether it will be in a compound with the Cl- or the NO3-. When potassium iodide is added I think that the Fe3+ will be reduced and the iodide will be oxidised to iodine but again I'm not sure how to put all this together.

Sounds good to me. If you write inic equations you don't need to worry about where to place the nitrate and chloride ions.

To determine the products of the reactions, we need to consider the individual reactions that occur when dilute nitric acid and potassium iodide are added to the solution obtained after dissolving iron metal in dilute hydrochloric acid (HCl).

Let's break down the reactions step by step:

1. Dissolving iron metal in dilute aqueous HCl:
Fe + 2HCl --> FeCl2 + H2

This reaction occurs because iron metal (Fe) reacts with hydrochloric acid (HCl) to form iron(II) chloride (FeCl2) and hydrogen gas (H2).

2. Warming the resulting solution with dilute nitric acid:
As you correctly mentioned, the iron(II) ions (Fe2+) present in the solution can be oxidized to iron(III) ions (Fe3+) by the nitric acid (HNO3). The addition of nitric acid is necessary to oxidize Fe2+ to Fe3+.

The reaction can be represented as:
2FeCl2 + 4HNO3 --> 2Fe(NO3)3 + 4HCl

This reaction shows that iron(II) chloride (FeCl2) reacts with nitric acid (HNO3) to form iron(III) nitrate (Fe(NO3)3) and hydrochloric acid (HCl).

3. Treatment with aqueous potassium iodide:
When aqueous potassium iodide (KI) is added to the resulting solution containing iron(III) ions (Fe3+), it will undergo a redox reaction. Iron(III) ions will be reduced to iron(II) ions, and iodide ions (I-) will be oxidized to iodine (I2).

The reaction can be represented as:
2Fe(NO3)3 + 6KI --> 2FeI3 + 6KNO3

This reaction shows that iron(III) nitrate (Fe(NO3)3) reacts with potassium iodide (KI) to form iron(III) iodide (FeI3) and potassium nitrate (KNO3).

Overall, the complete reaction can be summarized as follows:
Fe + 2HCl + 4HNO3 + 6KI --> 2FeI3 + 4HCl + 6KNO3 + H2

Remember to balance the equations and pay attention to the states of the species involved (solid, liquid, gas, aqueous).