The **molar volume** is the volume occupied by one mole of a substance (chemical element or chemical compound) at a given temperature and pressure.

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### Molar Volume

**How to find the molar volume of a gas using the ideal gas law?**

The most common molar volume is the molar volume of an ideal gas at standard temperature and pressure (273 K and 1.00 atm).

The molar volume is the volume occupied by 1 mol of a gas at standard temperature and pressure (STP). It can be calculated using PV = nRT.

### Gas volumes from moles and grams

_{2}

= number of moles of CO_{2} × 22.4 L

= 5 × 22.4

= 112 L

b) Volume of CO_{2}

= number of moles of CO_{2} × 22.4 L

= 0.5 × 22.4

= 11.2 L

**How to convert from grams to moles to liters?**

The following video shows an example of grams to moles to liters conversion.

It shows how to convert grams of a substance to liters at STP.

Example:

What is the volume of 5.643g of CO_{s} at STP?

### Moles from Gas Volume

**How to convert from liters to moles?**

The following video shows an example of liters to moles conversion. It shows how to convert litres of a gas at STP into moles

Example:

How many moles are there in 60.2L of CO_{s} at STP?

### Gas volumes from equations

**Step 1: **Write a balanced equation for the reaction.

2H_{2} (*g*) + O_{2} (*g*) → 2H_{2}O (*l*)

**Step 2:** Calculate the volume.

From the equation, 2 volumes of hydrogen react with 1 of oxygen or

2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen.

The volume of hydrogen that will react is 44.8 liters.**Step 1: **Write a balanced equation for the reaction.

S (*s*) + O_{2} (*g*) → SO_{2} (*g*)

**Step 2: **Get the number of moles from the grams.

32 g of sulfur atoms = 1 mole of sulfur atoms

So, 1 g = 1 ÷ 32 mole or 0.03125 moles of sulfur atoms

1 mole of sulfur atoms gives 1 mole of sulfur dioxide molecules

So, 0.03125 moles of sulfur atoms gives 0.03125 moles of sulfur dioxide

**Step 3:** Get the volume.

1 mole of sulfur dioxide molecules has a volume of 22.4 at STP

So, 0,03125 moles has a volume of 0.03125 × 22.4 = 0.7 liters at STP

So, 0.7 liters of sulfur dioxide are produced.

**How to solve equation stoichiometry questions with gases?**

Examples and practice problems of solving equation stoichiometry questions with gases. We calculate moles with 22.4 L at STP, and use molar mass (molecular weight) and mole ratios to figure out how many products or reactants we have.

Example:

How many grams of H_{2}O will be produced by 58.2L of CH_{4} at STP? Assume an excess of O_{2}.
Examples and practice problems of solving equation stoichiometry questions with gases. We calculate moles with the Ideal Gas Law, because the conditions are not at STP, and use molar mass (molecular weight) and mole ratios to figure out how many products or reactants we have.

Example:

If 85.0 g of NaN_{3} decomposes at 75°C and 2.30 atm, what volume of N_{2} will be made?

You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.

There are two standards, commonly used in schools:

- STP (standard temperature and pressure) which is 0 ºC and 1 atmosphere.
- RTP (room temperature and pressure) which is 20 ºC and 1 atmosphere.

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**Avogadro’s Law** states that:

1 mole of every gas occupies the same volume, at the same temperature and pressure.

At STP (standard temperature and pressure), this volume is 22.4 liters

At RTP (room temperature and pressure), this volume is 24 dm^{3} (liters)

We can also say:

**The molar volume of a gas is 22.4 liters at STP (standard temperature and pressure). **

**The molar volume of gas is 24 dm ^{3} at RTP (room temperature and pressure). **

The most common molar volume is the molar volume of an ideal gas at standard temperature and pressure (273 K and 1.00 atm).

The molar volume is the volume occupied by 1 mol of a gas at standard temperature and pressure (STP). It can be calculated using PV = nRT.

*Example: *

Calculate the volume of carbon dioxide gas, CO_{2}, occupied by (a) 5 moles and (b) 0.5 moles of the gas occupied at STP.

*Solution:*

= number of moles of CO

= 5 × 22.4

= 112 L

b) Volume of CO

= number of moles of CO

= 0.5 × 22.4

= 11.2 L

The following video shows an example of grams to moles to liters conversion.

It shows how to convert grams of a substance to liters at STP.

Example:

What is the volume of 5.643g of CO

*Example: *

Calculate the number of moles of ammonia gas, NH_{3}, in a volume of 80 L of the gas measured at STP.

*Solution: *

Volume of gas = number of moles × 22.414 L/mol

The following video shows an example of liters to moles conversion. It shows how to convert litres of a gas at STP into moles

Example:

How many moles are there in 60.2L of CO

From the equation for a reaction, we can tell how many moles of a gas take part. Using Avogadro’s Law, we can also work out its volume.

*Example: *

What volume of hydrogen will react with 22.4 liters of oxygen to form water? (All volumes are measured at STP)

*Solution: *

2H

From the equation, 2 volumes of hydrogen react with 1 of oxygen or

2 × 22.4 liters of hydrogen react with 22.4 liters of oxygen.

The volume of hydrogen that will react is 44.8 liters.

*Example: *

When sulfur burns in air it forms sulfur dioxide. What volume of this gas is produced when 1 g of sulfur burns? (A_{r} : S = 32) (All volumes are measured at STP)

*Solution: *

S (

32 g of sulfur atoms = 1 mole of sulfur atoms

So, 1 g = 1 ÷ 32 mole or 0.03125 moles of sulfur atoms

1 mole of sulfur atoms gives 1 mole of sulfur dioxide molecules

So, 0.03125 moles of sulfur atoms gives 0.03125 moles of sulfur dioxide

1 mole of sulfur dioxide molecules has a volume of 22.4 at STP

So, 0,03125 moles has a volume of 0.03125 × 22.4 = 0.7 liters at STP

So, 0.7 liters of sulfur dioxide are produced.

Examples and practice problems of solving equation stoichiometry questions with gases. We calculate moles with 22.4 L at STP, and use molar mass (molecular weight) and mole ratios to figure out how many products or reactants we have.

Example:

How many grams of H

Example:

If 85.0 g of NaN

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