1. Aqueous solutions of chromium(III) iodide and sodium hydroxide react to give a chromium(III) hydroxide precipitate and aqueous sodium iodide.
2. Aqueous solutions of iron(II) sulfate and barium nitrate react to give a barium sulfate precipitate and aqueous iron(II) nitrate.
Express your answer as a chemical equation. Identify all of the phases in your answer. Enter noreaction if there is no reaction.
1- The original equation of the reaction is : CrI3 + NaOH → Cr(OH)3 + NaI but this is not a balanced equation so we have to balance it : -first, we have 3 I atoms on the left side and 1 I atom on the right side so to makes the no.of atoms on both sides equal we put 3NaI instead of NaI CrI3 + NaOH → Cr(OH)3 + 3NaI - then no.of Na atoms on the left side is 1 and on the right side is 3 so, we put 3NaOH instead of NaOH CrI3 + 3NaOH → Cr(OH)3 + 3NaI now we have : Cr atoms = 1 on both sides I atoms = 3 on both sides Na atoms = 3 on both sides O&H atoms = 3 on both sides So the final balanced equation of our reaction is: CrI3(aq) + 3NaOH(aq) → Cr(OH)3(s) + 3NaI(aq)
2 - the chemical equation of the second reaction is : FeSO4 + Ba(NO3)2 → BaSO4 + Fe(NO3)2 let's see if it is the balanced equation or not? we have: no of Fe atom on both sides = 1 no of (SO4) atom on both sides are equal no of Ba atom on both sides = 1 no of (NO3) group on both sides= 2 so this is our balanced equation, so the final equation for this reaction is: FeSO4(aq) + Ba(NO3)2(aq) →BaSO4(s) + Fe(NO3)2(aq)
