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bekas [8.4K]
2 years ago
14

For the reaction between aqueous silver nitrate and aqueous sodium chloride, write each of the following. The products of the re

action are aqueous sodium nitrate and solid silver chloride.
Complete equation
Complete ionic equation
Net ionic equation
Chemistry
1 answer:
vova2212 [387]2 years ago
4 0

Answer:

A balanced ionic equation shows the reacting ions in a chemical reaction. These equations can be used to represent what happens in precipitation reactions or displacement reactions.

Precipitation reactions

In a typical precipitation reaction, two soluble reactants form an insoluble product and a soluble product.

For example, silver nitrate solution reacts with sodium chloride solution. Insoluble solid silver chloride and sodium nitrate solution form:

AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)

The Na+ ions and NO3- ions remain separate in the sodium nitrate solution and do not form a precipitate. Ions that remain essentially unchanged during a reaction are called spectator ions.This means these can be ignored when writing the ionic equation. Only how the solid silver chloride forms is needed to be shown:

Ag+(aq) + Cl-(aq) → AgCl(s)

In a balanced ionic equation:

the number of positive and negative charges is the same

the numbers of atoms of each element on the left and right are the same

Displacement reactions

Displacement reactions take place when a reactive element displaces a less reactive element from one of its compounds.

A common type of displacement reaction takes place when a reactive metal reacts with the salt of a less reactive metal. For example, copper reacts with silver nitrate solution to produce silver and copper(II) nitrate solution:

2AgNO3(aq) + Cu(s) → 2Ag(s) + Cu(NO3)2(aq)

In this reaction, the NO3- ions remain in the solution and do not react - they are the spectator ions in this reaction. So, they can be removed from the ionic equation:

2Ag+(aq) + Cu(s) → 2Ag(s) + Cu2+(aq)

Question

Explain why this ionic equation is balanced:

Ba2+(aq) + SO42-(aq) → BaSO4(s)

Hide answer

There are the same numbers of atoms of each element on both sides of the equation. The total charge on both sides is also the same (zero).

Question

Balance this ionic equation, which represents the formation of a silver carbonate precipitate:

Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Hide answer

2Ag+(aq) + CO32-(aq) → Ag2CO3(s)

Question

Balance this ionic equation, which represents the displacement of iodine from iodide ions by chlorine:

Cl2(aq) + I-(aq) → I2(aq) + Cl-(aq)

Hide answer

Cl2(aq) + 2I-(aq) → I2(aq) + 2Cl-(aq

Explanation:

this will help, I used this for my work x

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Write a complete, balanced chemical equation where tin metal reacts with aqueous hydrochloric acid to produce tin(II) chloride a
AleksAgata [21]

Answer:

1. The balanced equation is given below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

2a. H is oxidized.

2b. Sn is reduced.

Explanation:

1. Balanced equation for the reaction between tin (Sn) metal and aqueous hydrochloric acid (HCl) to produce tin(II) chloride (SnCl₂) and hydrogen gas (H₂).

This is illustrated below:

Sn (s) + HCl (aq) –> SnCl₂ (aq) + H₂ (g)

There are 2 atoms of Cl on the right side and 1 atom on the left side. It can be balance by putting 2 in front of HCl as shown below:

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

Now, the equation is balanced

2. Determination of the element that is oxidize and reduced.

This can be obtained as follow:

We shall determine the change in oxidation number of each element.

NOTE:

a. The oxidation number of H is always +1 except in hydrides where it is –1.

b. The oxidation state of Cl is always –1.

Sn (s) + 2HCl (aq) –> SnCl₂ (aq) + H₂ (g)

For Tin (Sn):

Sn = 0

SnCl₂ = 0

Sn + 2Cl = 0

Cl = – 1

Sn + 2(–1) = 0

Sn – 2 = 0

Collect like terms

Sn = 0 + 2

Sn = +2

Therefore, the oxidation number of Tin (Sn) changes from 0 to +2

For H:

H = +1

H₂ = 0

The oxidation number of H changes from +1 to 0

For Cl:

Cl is always –1. Therefore no change.

Summary:

Element >>Change in oxidation number

Sn >>>>>>>From 0 to +2

H >>>>>>>>From +1 to 0

Cl >>>>>>>No change

Therefore,

Sn is reduced since its oxidation number increased from 0 to +2.

H is oxidized since it oxidation number reduced from +1 to 0

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