Since it is not possible to calculate the weight of particles individually, a collection of such particles called **mole** is taken for all practical purposes. Avogadro discovered that under standard conditions of temperature and pressure, (1atm and 273K) a sample of gas occupies a volume of 22.4 L.

It was discovered that the number of atoms present in 12g of carbon is 6.023 x 10^{2}^{3} atoms. This is referred to as Avogadro number.

A mole of a gas is the amount of a substance containing 6.023 x 10^{2}^{3} particles.

Avogadro number is the number of atoms present in C^{1}^{2} isotope = 12g of C.

One mole of any gas at STP will have a volume of 22.4 L.

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Volumes of gases change with changes of temperature and pressure. Therefore **standard temperature and pressure **(abbreviated STP) have been chosen of comparing volumes of gases. Standard temperature is 0^{o}C and standard pressure is 1 atm or 760mm. Volumes of gases are corrected to STP prior to most stoichiometric calculations by performing a combined Boyle's law-Charles law calculation.

The number of moles of gas is required to perform stoichiometric calculation using the**mole concept**. It has been found experimentally that one mole of any ideal gas at STP occupies 22.4 liters. Thus 1.00 mole H_{2} (2.02 grams) at STP occupies 22.4 liters; 0.500 mole H_{2} (1.01 gram) at STP occupies 11.2 liters.

### Example Problem

The actual temperature and pressure at which we compare two or more gases do not matter. It is convenient however to select one temperature and one pressure as standard and chemists have chosen 1atm as the standard pressure and 0^{o}C as the standard temperature. These conditions are called **Standard temperature and pressure **(STP).

**Avogadro law:** Equal volumes of gases measured at the same temperature and pressure contain equal number of molecules.

Avogadro law allows us to write a gas that is valid not only for any pressure,volume and temperature but also for any quantity of gas. This law called the ideal gas law expressed as

**PV = nRT**

Where,

P = Pressure of the gas in atmosphere (atm)

V = Volume of gas in liters (L)

n = Amount of gas in moles (mol)

T = Temperature of gas in kelvin (K)

R = Ideal gas constant

**Problem 1: In the equation ****2C**_{2}H_{6} + 7O_{2} →** 4CO**_{2} + 6H_{2}O, how many molecules of C_{2}H_{6} are needed to produce 27.0 grams of H_{2}O?__Solution:__

The solution map for this problem is given below.

The problem is solved by factor label method (FLV)

**Problem 2: In the equation 4NH**_{3}(g) + 5O_{2(g)} →** 4NO**_{(g)} + 6H_{2}O(l) how many liters of NH_{3}(g) at STP are needed to react with 200 liters of O_{2}(g) at STP?__Solution:__

To solve this problem we need to convert between volume and numbers of moles. Our solution map is

The solution is given below

Volume problems are particularly easy to solve because the factor 22.4 always cancels in the calculations.

**Problem 3: How many grams of O**_{2} are required to burn completely 114g of octane.__Solution:__

The combustion reaction is written as follows

Molecular weight of octane = 114g

Number of moles of octane = 114 / 114 = 1 mole.

According to the equation one mole of octane reacts with 25 / 2 moles of O_{2}.

Therefore,

Weight of O_{2} = Number of moles x Molecular weight

= (25 / 2) x 32

= 400g

**Gram molar volume**

The volume occupied by one mole of a gas at STP is known as its gram molar volume.

1 mole of gas = 6.023 x 10^{23} molecules = 22.4 L in STP

Moles of a gas = Volume of the gas at STP in L / 22.4 L

**Problem 4: Calculate the volume of O**_{2} at STP produced by heating 12,25g of KClO_{3}**Solution:**

Molecular weight of KClO_{3} = 122.5g

**2KClO**_{3} →** 2KCl + 3O**_{2}

Number of moles of KClO_{3} taken = 12.25 / 122.5 = 1/10 = 0.1mole

According to the equation

2 moles of KClO_{3} = 3 moles of O_{2} = 3 x 22.4 L of O_{2} at STP

0.1 mole of KClO_{3} = {(3 x 0.1) / 2} moles of O_{2}

= {(3 x 0.1) x 22.4 / 2} L of O_{2} at STP

= 3.36 L of O_{2} at STP

**Problem 5: Calculate the amount of water formed when a mixture of 1g H**_{2} and 2g O_{2} is sparked.**Solution**:

1 gm H_{2} = 1 / 2 mole H_{2}; 2gm O_{2} = 2 / 32 mole O_{2} or 1 / 16 mole O_{2}

Now the equation for the formation of water is

**2H**_{2} + O_{2} →** 2H**_{2}O

(2mole) (1mole) (2mole)

**Problem 6: 56ml of NH**_{3} at STP is passed over heated cupric oxide to produce nitrogen. How many grams of magnesium nitride can be obtained from the liberated N_{2}.__Solution:__

The reaction are

The number of moles of gas is required to perform stoichiometric calculation using the

**Example 1:** A sample of hydrogen gas has a volume of 1.10 liters at -40^{o}C and 0.520atm. What will be its volume at STP.

**Solution:**Convert the required temperature to K.

K = C + 273 = -40 + 273 = 273K (T

K = C + 273 = 0 + 273 = 273K (T

P

P

Using the combined gas equation

**P _{1}V_{1} / T_{1} = P_{2}V_{2} / T_{2}**

Then V_{2} (Volume of H_{2} at STP) = (1.10 liters) {(0.520atm / 1.00atm) (273K / 273K)}

**Ans: V _{2} = 0.670 liters**

**Example 2:** Calculate the number of moles of hydrogen collected in the above example problem.

**Solution:**

The volume of H_{2} at STP is ** **0.670 liters. Determine the number of moles of H_{2} by dividing the volume by the molar volume of an ideal gas.

Number of moles of H_{2} = {0.670 liters / (22.4 liter / mole)}

**Ans: Number of moles of H _{2} = 0.0299 moles of H_{2}**

The actual temperature and pressure at which we compare two or more gases do not matter. It is convenient however to select one temperature and one pressure as standard and chemists have chosen 1atm as the standard pressure and 0

Avogadro law allows us to write a gas that is valid not only for any pressure,volume and temperature but also for any quantity of gas. This law called the ideal gas law expressed as

Where,

P = Pressure of the gas in atmosphere (atm)

V = Volume of gas in liters (L)

n = Amount of gas in moles (mol)

R = Ideal gas constant

The solution map for this problem is given below.

The problem is solved by factor label method (FLV)

The solution is given below

Volume problems are particularly easy to solve because the factor 22.4 always cancels in the calculations.

Number of moles of octane = 114 / 114 = 1 mole.

According to the equation one mole of octane reacts with 25 / 2 moles of O

Therefore,

Weight of O

= (25 / 2) x 32

= 400g

The volume occupied by one mole of a gas at STP is known as its gram molar volume.

1 mole of gas = 6.023 x 10

Moles of a gas = Volume of the gas at STP in L / 22.4 L

Molecular weight of KClO

Number of moles of KClO

According to the equation

2 moles of KClO

0.1 mole of KClO

= {(3 x 0.1) x 22.4 / 2} L of O

= 3.36 L of O

1 gm H

Now the equation for the formation of water is

(2mole) (1mole) (2mole)

From the above equation 1 mole H_{2} requires 1 / 2 mole oxygen therefore 1 / 2 mole H_{2} requires 1 / 4 mole O_{2} for complete conversion to water. Since number of moles of O_{2} available is much less therefore all of the oxygen will be converted to water but not all of the hydrogen.

Therefore from the equation 1 / 16 mole O_{2} will produce 2 / 16 mole H_{2}O, that is

(2 x 18) / 16

(or)**2.25 gm H**_{2}O

Therefore from the equation 1 / 16 mole O

(2 x 18) / 16

(or)

The reaction are

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