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Bess [88]
3 years ago
7

In a titration experiment, how many moles of naoh will be required to completely neutralize 1 mole of nitric acid?

Chemistry
2 answers:
katrin [286]3 years ago
4 0

Answer:

One mole

Explanation:

The balanced chemical equation is

<u>1</u>NaOH + <u>1</u>HNO₃ ⟶ NaNO₃ + H₂O

<u>1</u> mol       <u>1</u> mol

The coefficients in front of the formulas tell you the amount of something that reacts with an equivalent amount of something else.

In this reaction, 1 mol NaOH reacts with 1 mol HNO₃.

Doss [256]3 years ago
4 0

The number of moles of NaOH required to neutralize {\text{HN}}{{\text{O}}_{\text{3}}} of is \boxed{{\text{1}}\;{\text{mole}}}.

Further explanation:

<u>Neutralization reaction: </u>

It is the reaction that occurs between an acid and a base in order to form salt and water. It is named so as it neutralizes the excess amount of hydrogen or hydroxide ions present in the solution. It is used to decrease the acidity in the stomach, wastewater treatment, antacid tablets, and to control the pH of soil.

Titration:

It is a laboratory technique to determine the concentration of an unknown solution with the help of a known concentration. The known solution (titrant) is added from the burette to a known quantity of unknown solution (analyte).

{\text{HN}}{{\text{O}}_3} is a strong monoprotic acid and {\text{NaOH}} is also a strong base. The titration between {\text{HN}}{{\text{O}}_3} and {\text{NaOH}} would result in the following neutralization reaction. The neutralization reaction between {\text{HN}}{{\text{O}}_3} and {\text{NaOH}} is as follows:

 {\text{HN}}{{\text{O}}_3} + {\text{NaOH}} \to {\text{NaN}}{{\text{O}}_3} + {{\text{H}}_2}{\text{O}}

Since both acids and bases are strong and therefore complete neutralization occurs. The salt and water are formed as products. So {\mathbf{1}}\;{\mathbf{mole}} of {\text{HN}}{{\text{O}}_3} will be neutralized by {\mathbf{1}}\;{\mathbf{mole}} of {\text{NaOH}}.

Learn more:

1. Calculate volume of   required to titrate at equivalence point: brainly.com/question/10691070

2. What will be pH when monoprotic acid and base is titrated: brainly.com/question/3168961

Answer details:

Grade: High school

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: Neutralization reaction, HNO3, NaOH, monoprotic, strong, acid, base and titration.

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2 years ago
If 125g of KClO3 is heated, what is the total mass of the products?​
Andrej [43]

Given parameters:

Mass of KClO₃  = 125g

Unknown:

Total mass of the products = ?

When  KClO₃ is heated, it thermally decomposes to KCl and O₂ according to the chemical equation below;

               2KClO₃  →  2KCl + 3O₂

All chemical equations obeys the law of conservation of matter and with this regard, we know that the amount of reactants used is the same as that of the product.

The total mass of the products must give us 125g according to this law of conservation of matter.

Now to find the masses of each product,

  1. Find the number of moles of the given reactant:

     Number of moles  = \frac{mass}{molar mass}

  molar mass of  KClO₃  = 39 + 35.5 + 3(16)  = 122.5g/mol

    So number of moles of KClO₃ = \frac{125}{122.5}  = 1.02moles

    2. Now, using this number of moles, find the number of moles of the products using this value;

   2 moles of KClO₃ produced 2 moles of KCl

  1.02 moles of KClO₃ will also produce 1.02moles of KCl

   2 moles of KClO₃ produced 3 moles of O₂

   1.02 moles of KClO₃ will produce   \frac{1.02 x 3} {2} mole = 1.53 moles of O₂

   3. Now find the masses of each product;

Mass  = number of moles x molar mass

  molar mass of KCl  = 39 + 35.5 = 74.5g/mol

  molar mass of O₂  = 16 x 2  = 32g/mol

  Mass of KCl  = 74.5 x 1.02  = 75.99g

  Mass of O₂  = 32 x 1.53 = 48.96g

Total mass of products = mass of KCl + Mass of O₂ = 75.99g + 48.96g

                                        = 124.95g

This value is approximately the same as that of mass of  KClO₃

 

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3 years ago
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dezoksy [38]

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4 0
3 years ago
Collected data consists of:
san4es73 [151]

The number of moles of the magnesium (mg) is 0.00067 mol.

The number of moles of hydrogen gas is 0.0008 mol.

The volume of 1 more hydrogen gas (mL) at STP is 22.4 L.

<h3>Number of moles of the magnesium (mg)</h3>

The number of moles of the magnesium (mg) is calculated as follows;

number of moles = reacting mass / molar mass

molar mass of magnesium (mg) = 24 g/mol

number of moles = 0.016 g / 24 g/mol = 0.00067 mol.

<h3>Number of moles of hydrogen gas</h3>

PV = nRT

n = PV/RT

Apply Boyle's law to determine the change in volume.

P1V1 = P2V2

V2 = (P1V1)/P2

V2 = (101.39 x 146)/(116.54)

V2 = 127.02 mL

Now determine the number of moles using the following value of ideal constant.

R = 8.314 LkPa/mol.K

n = (15.15 kPa x 0.127 L)/(8.314 x 290.95)

n = 0.0008

<h3>Volume of 1 mole of hydrogen gas at STP</h3>

V = nRT/P

V = (1 x 8.314 x 273) / (101.325)

V = 22.4 L

Learn more about number of moles here: brainly.com/question/13314627

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