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aleksandr82 [10.1K]
3 years ago
8

Titration Lab Sheet: Day 2 (Alternate)‼️‼️‼️

Chemistry
2 answers:
zepelin [54]3 years ago
8 0
Hey did you ever find the answers to this?
hammer [34]3 years ago
6 0

Answer:

See explanation below

Explanation:

As the problem states, this is an acid base titration, and both titrations are already saying that they were both neutralized. When an acid base titration is neutralized, means that it reach it's equivalence point. In this point, we can say that the moles of the acid are the same moles of the base. In other words the following:

n₁ = n₂   (1)

1 is the acid and 2 is the base.

You should note that the above expression is real when the mole ratio is 1:1. When it's not, we need to see the mole ratio and then, adjust the expression to that.

the moles can also be expressed as:

n = M * V

Replacing in the first expression we have:

M₁V₁ = M₂V₂  (2)

With this expression we can calculate either the volume or concentration of the compounds given. Let's do this by parts:

<u>Titration 1:</u>

In this case we have KOH and H₂SO₄, so the balanced reaction would be:

2KOH + H₂SO₄ -------> K₂SO₄ + 2H₂O

As you can see, we have 2 moles of KOH and 1 mole of the acid, so the mole ratio is 2:1, therefore, expression (2) becomes:

M₁V₁ = 2M₂V₂

From here, we solve for concentration of the acid (M₁)

M₁ = 2M₂V₂ / V₁

Replacing the given values we have:

M₁ = 2 * 25 * 0.15 / 15

<h2>M₁ = 0.5 M</h2><h2>This is the concentration of the acid.</h2>

Now, how can we fill the chart? Is easy, we just put the obtained values:

For the acid it would be:

Solution: H₂SO₄;   Molar ratio: 1;    Volume: 15 mL;  Concentration: 0.5 M

For the base:

Solution: KOH;   Molar ratio: 2;    Volume: 25 mL;    Concentration: 0.15 M

<u>Titration 2:</u>

In this case we do the same thing as before but with different data. First the equation:

HBr + NaOH --------> NaBr + H₂O

The equation is already balanced and we can see a mole ratio of 1:1, so we can use expression (2) and solve for concentration of the base instead:

M₁V₁ = M₂V₂

M₂ = M₁V₁ / V₂

M₂ = 30 * 0.250 / 20

<h2>M₂ = 0.375 M</h2><h2>This is the concentration of the base.</h2>

The chart can be filled the same way as in titration 1:

For the acid it would be:

Solution: HBr;   Molar ratio: 1;    Volume: 30 mL;  Concentration: 0.25 M

For the base:

Solution: NaOH;   Molar ratio: 1;    Volume: 20 mL;    Concentration: 0.375 M

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The balanced equation is:

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The balanced equation is:

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The balanced equation is:

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D. The products are: CO2 and H2O

The balanced equation is:

C5H12 + 8O2 —> 5CO2 + 6H2O

Explanation:

A. ____ —> Li2O

The reactants are Li and O2. Thus the equation is given below:

Li + O2 —> Li2O

Thus the equation is balanced as follow:

There are 2 atoms of O on the left side and 1 atom on the right side. It can be balance by putting 2 in front of Li2O as shown below:

Li + O2 —> 2Li2O

Now, we have 4 atoms of Li on the right side and 1 atom on the left. It can be balance by putting 4 in front of Li as shown below:

4Li + O2 —> 2Li2O

Now the equation is balanced

B. Mg(ClO3)2 —> __

The products are MgCl2 and O2

The equation is given below:

Mg(ClO3)2 —> MgCl2 + O2

The equation can be balance as follow:

There are 6 atoms of O on the left side and 2 atoms on the right side. It can be balance by putting 3 in front of O2 as shown below:

Mg(ClO3)2 —> MgCl2 + 3O2

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The equation is given below:

HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

The equation is balanced as follow:

There are 2 atoms of NO3 on the right side and 1 atom on the left. It can be balance by putting 2 in front of HNO3 as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + H2O

There are a total of 4 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 2 in front of H2O as shown below:

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

Now the equation is balanced

D. C5H12 + O2 —>__

The products are: CO2 and H2O

The equation is given below:

C5H12 + O2 —> CO2 + H2O

The equation can be balance as follow:

There are 5 atoms of C on the left side and 1 atom on the right side. It can be balance by putting 5 in front of CO2 as shown below:

C5H12 + O2 —> 5CO2 + H2O

There are 12 atoms of H on the left side and 2 atoms on the right side. It can be balance by putting 6 in front of H2O as shown below:

C5H12 + O2 —> 5CO2 + 6H2O

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