Some Basic Concepts Of ChemistryClass 11 Chemistry Notes

DEVELOPMENT OF CHEMISTRY

Chemistry as we know it today is a relatively modern science, but its roots are ancient. Early chemical knowledge developed as a result of the search for two mythical substances:

1. Philosopher's stone (Paras): A substance believed to convert common metals like iron and copper into gold.
2. 'Elixir of life': A potion that would grant immortality.

This early form of chemistry was known as Alchemy.

In ancient India, chemistry, called Rasayan Shastra, Rastantra, Ras Kriya, or Rasvidya, was highly advanced. It included practices like metallurgy, medicine, and the creation of cosmetics, glass, and dyes.

Key Developments in Ancient India

• Metallurgy: Archaeological evidence from Mohenjodaro and Harappa shows the use of baked bricks and glazed pottery. Harappans were skilled in melting and forging metals like lead, silver, gold, and copper. They even created alloys by mixing copper with tin and arsenic to improve its hardness.
• Chemical Processes: The production of pottery, where materials were mixed, molded, and heated, can be seen as one of the earliest chemical processes. Kautilya's Arthashastra describes the production of salt from the sea.
• Atomic Theory: Around 600 BCE, Acharya Kanda (originally Kashyap) proposed an atomic theory. He described matter as being made of tiny, indivisible, eternal particles called 'Paramãnu' (similar to atoms). He suggested that these particles could combine in pairs or triplets, a concept that predates John Dalton's theory by about 2500 years.
• Medicine and Nanotechnology: The Charaka Samhita, an ancient Ayurvedic text, discusses the concept of reducing the particle size of metals for medicinal purposes. The use of bhasma (metallic preparations) in treatments is an early form of nanotechnology, as modern science has shown that these contain nanoparticles of metals.
• Other Discoveries:
  • Nagarjuna, a renowned chemist and metallurgist, detailed methods for extracting metals like gold and silver in his work Rasratnakar.
  • Chakrapani is credited with discovering mercury sulphide and inventing soap using mustard oil and alkalies.
  • Texts like the Rasopanishada describe the preparation of gunpowder.

Modern chemistry began to take shape in 18th-century Europe and was introduced to India in the latter half of the 19th century by European scientists.

IMPORTANCE OF CHEMISTRY

Chemistry is often called the central science because it connects with other branches like physics, biology, and geology. Its principles are essential in understanding everything from weather patterns to the functioning of our brains.

Chemistry plays a vital role in:

• National Economy: Chemical industries manufacture essential products like fertilizers, acids, salts, polymers, drugs, soaps, and detergents, contributing significantly to a nation's economy and creating jobs.
• Food and Agriculture: It has helped improve food production through the development of fertilizers, pesticides, and insecticides.
• Health and Medicine: Chemistry is crucial for developing life-saving drugs. For example:
  • Cisplatin and taxol are effective in cancer therapy.
  • AZT (Azidothymidine) is used to help AIDS patients.
• Improving Quality of Life: It has enabled the creation of new materials with specific properties, such as superconducting ceramics, conducting polymers, and optical fibres.
• Environmental Protection: Chemists are developing safer alternatives to harmful substances. For instance, environmentally friendly refrigerants have been synthesized to replace CFCs (chlorofluorocarbons), which deplete the ozone layer.

NATURE OF MATTER

Matter is anything that has mass and occupies space. Everything around us, including books, water, air, and living beings, is made of matter.

States of Matter

Matter exists in three physical states: solid, liquid, and gas. The state of a substance depends on how its constituent particles (atoms or molecules) are arranged.

• Solids:
  • Particles are packed tightly in an orderly fashion.
  • They have very little freedom of movement.
  • Have a definite volume and a definite shape.
• Liquids:
  • Particles are close to each other but can move around.
  • Have a definite volume but no definite shape; they take the shape of their container.
• Gases:
  • Particles are far apart from each other.
  • Their movement is easy and fast.
  • Have neither a definite volume nor a definite shape; they completely fill the container they are in.

These three states are interconvertible by changing temperature and pressure. $\text{Solid} \underset{\text{cool}}{\stackrel{\text{heat}}{\rightleftharpoons}} \text{Liquid} \underset{\text{cool}}{\stackrel{\text{heat}}{\rightleftharpoons}} \text{Gas}$

Classification of Matter

At a macroscopic (bulk) level, matter is classified into two main categories: mixtures and pure substances.

Mixtures

A mixture contains two or more pure substances (called components) in any ratio. Its composition is variable.

• Homogeneous Mixture: The components mix completely, and the composition is uniform throughout. You cannot see the individual components.
  • Examples: Sugar dissolved in water, air.
• Heterogeneous Mixture: The composition is not uniform, and the different components are often visible.
  • Examples: A mix of salt and sugar, grains mixed with dirt.

Pure Substances

A pure substance consists of only one type of particle and has a fixed composition.

• Elements: An element consists of only one type of atom. These atoms can exist individually or as molecules.
  • Examples: Copper (Cu), Sodium (Na), Hydrogen gas ($H_2$), Oxygen gas ($O_2$).
• Compounds: A compound is formed when atoms of two or more different elements combine in a fixed ratio. The properties of a compound are completely different from those of its constituent elements.
  • Example: Water ($H_2O$) is a compound made of hydrogen and oxygen. Hydrogen is a flammable gas, and oxygen supports combustion, but water is a liquid that extinguishes fire.

PROPERTIES OF MATTER AND THEIR MEASUREMENT

Every substance has unique properties that can be classified into two types.

• Physical Properties: These can be measured or observed without changing the substance's identity or composition.
  • Examples: Colour, odour, melting point, boiling point, density.
• Chemical Properties: These describe a substance's ability to undergo a chemical change. Observing them requires a chemical reaction.
  • Examples: Acidity, basicity, combustibility, reactivity with other substances.

Measurement of Physical Properties

Scientific investigations rely on quantitative measurements. A quantitative measurement is always expressed as a number followed by a unit. For example, "6 metres" (6 is the number, metres is the unit).

The International System of Units (SI)

To ensure uniformity, the scientific community uses the International System of Units (SI). It has seven base units for fundamental scientific quantities.

Other physical quantities like volume, speed, and density are derived units, meaning they are derived from these seven base units.

Mass and Weight

• Mass is the amount of matter in a substance. It is constant regardless of location. The SI unit of mass is the kilogram (kg). In labs, the gram (g) is more commonly used ($1 \text{ kg} = 1000 \text{ g}$).
• Weight is the force exerted by gravity on an object. It can change depending on the gravitational pull (e.g., your weight on the Moon is less than on Earth).

Volume

Volume is the amount of space occupied by a substance. The SI unit for volume is the cubic metre ($m^3$). In chemistry, smaller units are more common:

• Litre (L)
• Millilitre (mL)
• Cubic centimetre ($cm^3$)
• Cubic decimetre ($dm^3$)

Key Conversions:

• $1 \text{ L} = 1000 \text{ mL}$
• $1000 \text{ cm}^3 = 1 \text{ dm}^3$
• $1 \text{ L} = 1 \text{ dm}^3 = 1000 \text{ cm}^3$

Density

Density is the amount of mass per unit volume. It tells us how closely packed the particles of a substance are. $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$ The SI unit is $\text{kg m}^{-3}$, but it is often expressed in $\text{g cm}^{-3}$.

Temperature

There are three common scales for measuring temperature:

1. Celsius ($^{\circ}$C): Freezing point of water is $0^{\circ}\text{C}$, and boiling point is $100^{\circ}\text{C}$.
2. Fahrenheit ($^{\circ}$F): Freezing point of water is $32^{\circ}\text{F}$, and boiling point is $212^{\circ}\text{F}$.
3. Kelvin (K): This is the SI unit of temperature. Negative values are not possible on the Kelvin scale.

Temperature Conversion Formulas:

• Fahrenheit from Celsius: ${^{\circ}\text{F}} = \frac{9}{5}({^{\circ}\text{C}}) + 32$
• Kelvin from Celsius: $\text{K} = {^{\circ}\text{C}} + 273.15$

UNCERTAINTY IN MEASUREMENT

All experimental measurements have some degree of uncertainty due to limitations of the measuring instrument and the skill of the person taking the measurement. This uncertainty is managed using scientific notation and significant figures.

Scientific Notation

To handle very large or very small numbers, we use scientific notation. A number is expressed in the form $N \times 10^n$, where:

• $N$ is a number between 1.000... and 9.999...
• $n$ is an exponent (a positive or negative integer).

• The number 232.508 can be written as $2.32508 \times 10^2$. (Decimal moved 2 places to the left, so exponent is +2).
• The number 0.00016 can be written as $1.6 \times 10^{-4}$. (Decimal moved 4 places to the right, so exponent is -4).

Calculations with Scientific Notation:

• Multiplication and Division: Multiply/divide the $N$ terms and add/subtract the exponents.
  • $(5.6 \times 10^5) \times (6.9 \times 10^8) = (5.6 \times 6.9) \times 10^{5+8} = 38.64 \times 10^{13} = 3.864 \times 10^{14}$
• Addition and Subtraction: Adjust the numbers so they have the same exponent, then add or subtract the $N$ terms.
  • $(6.65 \times 10^4) + (8.95 \times 10^3) = (6.65 \times 10^4) + (0.895 \times 10^4) = (6.65 + 0.895) \times 10^4 = 7.545 \times 10^4$

Significant Figures

Significant figures are all the certain digits in a measurement plus one uncertain (estimated) digit. They indicate the precision of a measurement.

Rules for Determining Significant Figures:

1. All non-zero digits are significant. (e.g., 285 cm has 3 significant figures).
2. Zeros preceding the first non-zero digit are not significant. (e.g., 0.03 has 1 significant figure).
3. Zeros between two non-zero digits are significant. (e.g., 2.005 has 4 significant figures).
4. Zeros at the end of a number are significant if they are on the right side of a decimal point. (e.g., 0.200 g has 3 significant figures).
5. Terminal zeros without a decimal point are ambiguous. (e.g., 100 has only 1 significant figure, but 100. has 3). Using scientific notation avoids this:
   • $1 \times 10^2$ (1 significant figure)
   • $1.0 \times 10^2$ (2 significant figures)
   • $1.00 \times 10^2$ (3 significant figures)
6. Exact numbers (from counting objects like 2 balls, or defined quantities) have infinite significant figures.

Precision and Accuracy

• Precision refers to the closeness of various measurements for the same quantity.
• Accuracy is the agreement of a particular value to the true value of the result.

• Student A measures 1.95 g and 1.93 g. The results are precise (close to each other) but not accurate (far from the true value).
• Student B measures 1.94 g and 2.05 g. The results are neither precise nor accurate.
• Student C measures 2.01 g and 1.99 g. The results are both precise and accurate.

Calculations with Significant Figures

• Addition and Subtraction: The result cannot have more digits to the right of the decimal point than the number with the fewest decimal places.
  • $12.11 + 18.0 + 1.012 = 31.122$. Since 18.0 has only one decimal place, the answer is rounded to 31.1.
• Multiplication and Division: The result must have the same number of significant figures as the measurement with the fewest significant figures.
  • $2.5 \times 1.25 = 3.125$. Since 2.5 has two significant figures, the answer is rounded to 3.1.

Dimensional Analysis

Dimensional analysis (or the unit factor method) is a technique used to convert units from one system to another. It involves multiplying the given quantity by a unit factor, which is a fraction where the numerator and denominator are equivalent quantities with different units (e.g., $\frac{2.54 \text{ cm}}{1 \text{ in}}$).

Given

• Length = 3 in
• Conversion factor: 1 in = 2.54 cm

To Find

Length in cm

Formula

$\text{Length in cm} = \text{Length in in} \times \text{Unit Factor}$

Solution

The unit factor must have the desired unit (cm) in the numerator and the unit to be cancelled (in) in the denominator. $\text{Unit Factor} = \frac{2.54 \text{ cm}}{1 \text{ in}}$ $3 \text{ in} \times \frac{2.54 \text{ cm}}{1 \text{ in}} = 3 \times 2.54 \text{ cm} = 7.62 \text{ cm}$

Final Answer The length of the metal is 7.62 cm.

Given

• Volume = 2 L
• Conversion factors: $1 \text{ L} = 1000 \text{ cm}^3$ and $1 \text{ m} = 100 \text{ cm}$

To Find

Volume in $m^3$

Formula

$\text{Volume in } m^3 = \text{Volume in L} \times (\text{Conversion Factors})$

Solution

First, convert L to $cm^3$: $2 \text{ L} = 2 \times 1000 \text{ cm}^3 = 2000 \text{ cm}^3$ Next, create a unit factor to convert $cm^3$ to $m^3$. Since $1 \text{ m} = 100 \text{ cm}$, we cube both sides: $(1 \text{ m})^3 = (100 \text{ cm})^3$, which gives $1 \text{ m}^3 = 10^6 \text{ cm}^3$. The unit factor is $\frac{1 \text{ m}^3}{10^6 \text{ cm}^3}$. $2000 \text{ cm}^3 \times \frac{1 \text{ m}^3}{10^6 \text{ cm}^3} = \frac{2000}{10^6} \text{ m}^3 = 2 \times 10^{-3} \text{ m}^3$

Final Answer The volume of the milk is $2 \times 10^{-3} \text{ m}^3$.

LAWS OF CHEMICAL COMBINATIONS

The formation of compounds from elements is governed by five basic laws.

Law of Conservation of Mass

Proposed by Antoine Lavoisier in 1789.

Matter can neither be created nor destroyed.

This means that in any physical or chemical change, the total mass of the reactants is equal to the total mass of the products.

Law of Definite Proportions

Proposed by Joseph Proust.

A given compound always contains exactly the same proportion of elements by weight, regardless of its source or method of preparation.

For example, whether it's natural or synthetic, cupric carbonate always contains 51.35% copper, 38.91% oxygen, and 9.74% carbon by mass. This is also known as the Law of Definite Composition.

Law of Multiple Proportions

Proposed by John Dalton in 1803.

If two elements can combine to form more than one compound, the masses of one element that combine with a fixed mass of the other element are in the ratio of small whole numbers.

• In water: 2 g of hydrogen combines with 16 g of oxygen.
• In hydrogen peroxide: 2 g of hydrogen combines with 32 g of oxygen. The masses of oxygen (16 g and 32 g) that combine with a fixed mass of hydrogen (2 g) are in a simple ratio of $16:32$ or 1:2.

Gay Lussac's Law of Gaseous Volumes

Proposed by Gay Lussac in 1808.

When gases combine or are produced in a chemical reaction, they do so in a simple ratio by volume, provided all gases are at the same temperature and pressure.

For example, 100 mL of hydrogen gas combines with 50 mL of oxygen gas to produce 100 mL of water vapour. The ratio of volumes of hydrogen to oxygen is $100:50$ or 2:1.

Avogadro's Law

Proposed by Amedeo Avogadro in 1811.

Equal volumes of all gases at the same temperature and pressure should contain an equal number of molecules.

This law explained Gay Lussac's observations and made a crucial distinction between atoms and molecules. For example, it explains why 2 volumes of hydrogen and 1 volume of oxygen produce 2 volumes of water vapour, by proposing that hydrogen and oxygen exist as diatomic molecules ($H_2$ and $O_2$).

DALTON'S ATOMIC THEORY

In 1808, John Dalton proposed his atomic theory, which could explain the laws of chemical combination. Its main points are:

1. Matter consists of indivisible particles called atoms.
2. All atoms of a given element have identical properties and mass. Atoms of different elements have different masses.
3. Compounds are formed when atoms of different elements combine in a fixed, simple whole-number ratio.
4. Chemical reactions involve only the reorganization of atoms. Atoms are neither created nor destroyed in a chemical reaction.

ATOMIC AND MOLECULAR MASSES

Atomic Mass

The mass of an individual atom is extremely small. The modern system of atomic masses is based on the carbon-12 isotope ($^{12}$C) as the standard.

• One atomic mass unit (amu), now called a unified mass (u), is defined as a mass exactly equal to one-twelfth ($1/12$) the mass of one carbon-12 atom.
• $1 \text{ u} = 1.66056 \times 10^{-24} \text{ g}$

The atomic mass of an element is its mass relative to this standard. For example, the mass of a hydrogen atom is 1.0080 u.

Average Atomic Mass

Most elements exist naturally as a mixture of isotopes (atoms of the same element with different masses). The average atomic mass is a weighted average that accounts for the relative abundance of each isotope.

Molecular Mass

The molecular mass of a substance is the sum of the atomic masses of all the atoms in a molecule.

• Molecular mass of methane ($CH_4$) = (Atomic mass of C) + 4 $\times$ (Atomic mass of H) = $12.011 \text{ u} + 4 \times (1.008 \text{ u}) = 16.043 \text{ u}$.
• Molecular mass of water ($H_2O$) = 2 $\times$ (Atomic mass of H) + (Atomic mass of O) = $2 \times (1.008 \text{ u}) + 16.00 \text{ u} = 18.02 \text{ u}$.

Formula Mass

For ionic compounds like sodium chloride (NaCl) that exist as a crystal lattice rather than discrete molecules, we use formula mass. It is calculated the same way as molecular mass, by summing the atomic masses in the formula unit.

• Formula mass of NaCl = (Atomic mass of Na) + (Atomic mass of Cl) = $23.0 \text{ u} + 35.5 \text{ u} = 58.5 \text{ u}$.

Given

• Formula: $C_6H_{12}O_6$
• Atomic mass of C = 12.011 u
• Atomic mass of H = 1.008 u
• Atomic mass of O = 16.00 u

To Find

Molecular mass of glucose

Formula

$\text{Molecular Mass} = \sum (\text{Number of atoms of element} \times \text{Atomic mass of element})$

Solution

$\text{Molecular mass of } C_6H_{12}O_6 = 6(12.011 \text{ u}) + 12(1.008 \text{ u}) + 6(16.00 \text{ u})$ $= (72.066 \text{ u}) + (12.096 \text{ u}) + (96.00 \text{ u})$ $= 180.162 \text{ u}$

Final Answer The molecular mass of glucose is 180.162 u.

MOLE CONCEPT AND MOLAR MASSES

Atoms and molecules are so small that we deal with enormous numbers of them at once. The mole is a unit used to count these particles.

• The mole (mol) is the SI unit for the amount of a substance.
• One mole contains exactly $6.02214076 \times 10^{23}$ elementary entities (atoms, molecules, ions, etc.).
• This number is known as the Avogadro constant or Avogadro number ($N_A$).

So,

• 1 mole of hydrogen atoms = $6.022 \times 10^{23}$ hydrogen atoms.
• 1 mole of water molecules = $6.022 \times 10^{23}$ water molecules.

The molar mass of a substance is the mass of one mole of that substance in grams. It is numerically equal to the atomic, molecular, or formula mass in u.

• Atomic mass of Na = 23.0 u $\rightarrow$ Molar mass of Na = 23.0 g/mol
• Molecular mass of $H_2O$ = 18.02 u $\rightarrow$ Molar mass of $H_2O$ = 18.02 g/mol

PERCENTAGE COMPOSITION

Percentage composition tells us the percentage by mass of each element present in a compound. $\text{Mass \% of an element} = \frac{\text{Mass of that element in the compound}}{\text{Molar mass of the compound}} \times 100$

Given

• Formula: $C_2H_5OH$
• Molar mass of ethanol = $(2 \times 12.01) + (6 \times 1.008) + 16.00 = 46.068 \text{ g/mol}$
• Mass of Carbon in 1 mole = $2 \times 12.01 = 24.02 \text{ g}$
• Mass of Hydrogen in 1 mole = $6 \times 1.008 = 6.048 \text{ g}$
• Mass of Oxygen in 1 mole = $16.00 \text{ g}$

To Find

Mass per cent of C, H, and O.

Formula

$\text{Mass \%} = \frac{\text{Mass of element}}{\text{Molar mass}} \times 100$

Solution

$\text{Mass \% of Carbon} = \frac{24.02 \text{ g}}{46.068 \text{ g}} \times 100 = 52.14\%$ $\text{Mass \% of Hydrogen} = \frac{6.048 \text{ g}}{46.068 \text{ g}} \times 100 = 13.13\%$ $\text{Mass \% of Oxygen} = \frac{16.00 \text{ g}}{46.068 \text{ g}} \times 100 = 34.73\%$

Final Answer Ethanol contains 52.14% Carbon, 13.13% Hydrogen, and 34.73% Oxygen by mass.

Empirical Formula for Molecular Formula

• Empirical Formula: Represents the simplest whole-number ratio of atoms in a compound.
• Molecular Formula: Shows the exact number of atoms of each element in a molecule.

If the mass percent and molar mass are known, we can determine both formulas.

Given

• Mass % H = 4.07%
• Mass % C = 24.27%
• Mass % Cl = 71.65%
• Molar mass of compound = 98.96 g/mol

To Find

Empirical formula and molecular formula.

Solution

Step 1. Convert mass per cent to grams. Assume a 100 g sample. This gives us:

• Mass of H = 4.07 g
• Mass of C = 24.27 g
• Mass of Cl = 71.65 g

Step 2. Convert grams to moles for each element.

• Moles of H = $\frac{4.07 \text{ g}}{1.008 \text{ g/mol}} = 4.04 \text{ mol}$
• Moles of C = $\frac{24.27 \text{ g}}{12.01 \text{ g/mol}} = 2.021 \text{ mol}$
• Moles of Cl = $\frac{71.65 \text{ g}}{35.453 \text{ g/mol}} = 2.021 \text{ mol}$

Step 3. Divide by the smallest mole value to find the simplest ratio. The smallest value is 2.021.

• H: $\frac{4.04}{2.021} \approx 2$
• C: $\frac{2.021}{2.021} = 1$
• Cl: $\frac{2.021}{2.021} = 1$ The ratio is H:C:Cl = 2:1:1.

Step 4. Write the empirical formula. The simplest whole-number ratio gives the empirical formula: $CH_2Cl$.

Step 5. Find the molecular formula. (a) Calculate the empirical formula mass. Empirical formula mass of $CH_2Cl = 12.01 + (2 \times 1.008) + 35.453 = 49.48 \text{ g/mol}$.

(b) Find the ratio of the molar mass to the empirical formula mass. $n = \frac{\text{Molar mass}}{\text{Empirical formula mass}} = \frac{98.96 \text{ g/mol}}{49.48 \text{ g/mol}} \approx 2$

(c) Multiply the empirical formula by this ratio ($n$) to get the molecular formula. Molecular formula = $n \times (\text{Empirical Formula}) = 2 \times (CH_2Cl) = \bf{C_2H_4Cl_2}$.

Final Answer The empirical formula is $CH_2Cl$ and the molecular formula is $C_2H_4Cl_2$.

STOICHIOMETRY AND STOICHIOMETRIC CALCULATIONS

The word stoichiometry comes from the Greek words stoicheion (element) and metron (measure). It deals with the calculation of the masses (and sometimes volumes) of reactants and products in a chemical reaction.

A balanced chemical equation provides the quantitative relationships. Consider the combustion of methane: $\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{g})$ The numbers in front of the formulas are stoichiometric coefficients. This equation tells us:

• Mole ratio: 1 mole of $CH_4$ reacts with 2 moles of $O_2$ to produce 1 mole of $CO_2$ and 2 moles of $H_2O$.
• Mass ratio: 16 g of $CH_4$ reacts with 64 g of $O_2$ to produce 44 g of $CO_2$ and 36 g of $H_2O$.

Limiting Reagent

In many reactions, the reactants are not mixed in the exact stoichiometric ratio. The limiting reagent is the reactant that is completely consumed first in a chemical reaction. It determines, or "limits," the maximum amount of product that can be formed. The other reactant(s) are said to be in excess.

Reactions in Solutions

Many reactions occur in solutions. The amount of substance in a solution is expressed by its concentration. Common ways to express concentration include:

1. Mass per cent (w/w %): $\text{Mass per cent} = \frac{\text{Mass of solute}}{\text{Mass of solution}} \times 100$
2. Mole Fraction: The ratio of the number of moles of a particular component to the total number of moles in the solution. For a solution of substance A in substance B: $\text{Mole fraction of A} = \frac{n_A}{n_A + n_B}$
3. Molarity (M): The number of moles of solute dissolved in 1 litre of solution. $\text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$ [!note] Molarity is temperature-dependent because the volume of a solution can change with temperature.
4. Molality (m): The number of moles of solute present in 1 kg of solvent. $\text{Molality (m)} = \frac{\text{Number of moles of solute}}{\text{Mass of solvent in kg}}$ [!note] Molality is independent of temperature because it is based on mass, which does not change with temperature.

Given

• Mass of methane ($CH_4$) = 16 g
• Balanced equation: $CH_4(\text{g}) + 2O_2(\text{g}) \rightarrow CO_2(\text{g}) + 2H_2O(\text{g})$

To Find

Mass of water ($H_2O$) produced.

Solution

First, find the moles of methane. The molar mass of $CH_4$ is $12 + (4 \times 1) = 16 \text{ g/mol}$. $\text{Moles of } CH_4 = \frac{16 \text{ g}}{16 \text{ g/mol}} = 1 \text{ mol}$ From the balanced equation, 1 mole of $CH_4$ produces 2 moles of $H_2O$. Now, convert moles of $H_2O$ to grams. The molar mass of $H_2O$ is $(2 \times 1) + 16 = 18 \text{ g/mol}$. $\text{Mass of } H_2O = 2 \text{ mol} \times 18 \text{ g/mol} = 36 \text{ g}$

Final Answer 36 g of water is produced.

Given

• Mass of $N_2$ = 50.0 kg = $50000$ g
• Mass of $H_2$ = 10.0 kg = $10000$ g
• Balanced equation: $N_2(\text{g}) + 3H_2(\text{g}) \rightleftharpoons 2NH_3(\text{g})$

To Find

• Mass of $NH_3$ formed
• The limiting reagent

Solution

1. Calculate the moles of each reactant. Molar mass of $N_2$ = 28.0 g/mol. $\text{Moles of } N_2 = \frac{50000 \text{ g}}{28.0 \text{ g/mol}} = 1786 \text{ mol} \approx 1.786 \times 10^3 \text{ mol}$ Molar mass of $H_2$ = 2.016 g/mol. $\text{Moles of } H_2 = \frac{10000 \text{ g}}{2.016 \text{ g/mol}} = 4960 \text{ mol} \approx 4.96 \times 10^3 \text{ mol}$

2. Identify the limiting reagent. According to the balanced equation, 1 mole of $N_2$ requires 3 moles of $H_2$. Let's see how much $H_2$ is needed to react with all the $N_2$. $\text{Moles of } H_2 \text{ needed} = 1786 \text{ mol } N_2 \times \frac{3 \text{ mol } H_2}{1 \text{ mol } N_2} = 5358 \text{ mol } H_2$ We need 5358 moles of $H_2$, but we only have 4960 moles. Therefore, $H_2$ is the limiting reagent. The reaction will stop when all the $H_2$ is used up.

3. Calculate the amount of product formed. The calculation must be based on the limiting reagent ($H_2$). From the equation, 3 moles of $H_2$ produce 2 moles of $NH_3$. $\text{Moles of } NH_3 \text{ formed} = 4960 \text{ mol } H_2 \times \frac{2 \text{ mol } NH_3}{3 \text{ mol } H_2} = 3307 \text{ mol } NH_3 \approx 3.30 \times 10^3 \text{ mol}$

4. Convert moles of product to mass. Molar mass of $NH_3$ = 17.0 g/mol. $\text{Mass of } NH_3 = 3307 \text{ mol} \times 17.0 \text{ g/mol} = 56219 \text{ g} \approx 56.2 \text{ kg}$

Final Answer The amount of $NH_3$ formed is 56.2 kg. The limiting reagent is dihydrogen ($H_2$).

Given

• Mass of NaOH = 4 g
• Volume of solution = 250 mL = 0.250 L
• Molar mass of NaOH = 40 g/mol

To Find

Molarity (M) of the NaOH solution.

Formula

$\text{Molarity (M)} = \frac{\text{Number of moles of solute}}{\text{Volume of solution in litres}}$

Solution

First, calculate the number of moles of NaOH. $\text{Moles of NaOH} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{4 \text{ g}}{40 \text{ g/mol}} = 0.1 \text{ mol}$ Now, calculate the molarity. $M = \frac{0.1 \text{ mol}}{0.250 \text{ L}} = 0.4 \text{ mol/L} = 0.4 \text{ M}$

Final Answer The molarity of the NaOH solution is 0.4 M.