Identify and matching

Question

1. Identify all of the species present in the net ionic equation for the reaction of sodium sulfate and calcium chloride.

Ca²⁺

CaCl₂

SO₄^2-

NaCl

Cl^-

CaSO₄

Na₂SO₄

Na⁺

2.dentify all precipitates of the following reactions: ammonium hydroxide reacts with copper (II) nitratepotassium carbonate reacts with magnesium bromide sodium oxide reacts with aluminum sulfate

Na₂SO₄

Al₂O₃

Cu(OH)₂

MgO

MgCO₃

Cu(NH₄)₂

NH₄NO₃

KBr

3.Match each species from the following reaction with it's correct redox terminology.

Fe₂O₃ + SnBr₂ -----> FeBr₂ + SnO₂

1. The species that is reduced.
2. The species that is oxidized.
3. The oxidizing agent.
4. The reducing agent.

A. Fe²⁺.

B. SnBr₂

C. FeBr₂

D. Sn⁴⁺

E. SnO₂

F. Fe₂O₃

G.Sn²⁺

H. Fe³⁺

Answer 1

This is a longer video response, so I wanted to leave the answers here in the text for you:

#1: The net ionic equation only includes the precipitates and the ions that reacted to form them. The full reaction is:

Na₂SO₄ + CaCl₂ --> 2 NaCl + CaSO₄

Since calcium sulfate is the precipitate of this reaction, the net ionic equation would be:

SO₄^2- + Ca²⁺ --> CaSO₄

#2: Looking at each reaction, and using solubility rules, we can determine which product is a precipitate and which is aqueous.

2 NH₄OH + Cu(NO₃)₂ --> Cu(OH)₂ + 2 NH₄NO₃

Anything with nitrate and ammonium is aqueous, and most things with hydroxide are insoluble. Therefore, copper II hydroxide is the precipitate.

K₂CO₃ + MgBr₂ --> 2 KBr + MgCO₃

Group 1A and 7A ions are almost always soluble (with a few exceptions). Carbonates are almost always insoluble. Since K and Br are from groups 1 and 7 respectively, the precipitate is magnesium carbonate.

3 Na₂O + Al₂(SO₄)₃ --> 3 Na₂SO₄ + Al₂O₃

Again, anything from group 1 is soluble, making sodium sulfate soluble. Aluminum and oxide don't have a specific rule. The general rule is if there's no rule, then the ionic compound is insoluble. Therefore, aluminum oxide is the precipitate.

Precipitates: Cu(OH)₂, MgCO₃, and Al₂O₃

#3. The oxidation states of oxygen and bromine remain the same throughout the reaction. Therefore, we are concerned only about iron and tin. Iron starts as iron (III) and is reduced to iron (II). Tin starts as tin (II) and is oxidized to tin (IV). I explain the ideas behind redox more in the video response.

When it asks for the "species" being reduced or oxidized, it usually means the whole molecule / formula unit. So Fe₂O₃ was reduced, and SnBr₂ was oxidized.

The oxidizing agent is the thing that was responsible for oxidation occurring. Since oxidation is a loss of electrons, the oxidizing agent would be the thing that received the lost electrons. Likewise, the reducing agent is the thing responsible for reduction occurring. Since reduction involved gaining electrons, the reducing agent is the thing that supplies the electrons that are being gained.

Since iron (III) was reduced to iron (II), it received the lost electrons, so the iron (III) ion was the oxidizing agent.

Since tin (II) was oxidized to tin (IV), it gave up electrons in order for something else to receive them to be reduced. Therefore, the tin (II) ion was the reducing agent.

Notice that just the ion is the oxidizing agent, not the whole compound.

Species reduced: Fe₂O₃

Species Oxidized: SnBr₂

Oxidizing agent: Fe³⁺

Reducing agent: Sn²⁺