GCSE Chemistry - How to Find the Volume of a Gas #28
Summary
TLDRThis video explains how to calculate the volume of gases using a simple equation that multiplies the number of moles by 24, which works for all gases at room temperature and pressure. It covers examples like chlorine, oxygen, and water vapor, and discusses rearranging the equation to find moles from volume. The video also addresses finding gas volumes from chemical reactions using molar ratios, and highlights shortcuts when only converting between gas volumes. The method applies under standard conditions, emphasizing its use in exams.
Takeaways
- 📏 The equation for calculating the volume of a gas (in dm³) is simple: multiply the number of moles by 24.
- 🧪 The type of gas (e.g., chlorine, water vapor, oxygen) doesn't matter in this equation.
- 🔢 Example: 3.5 moles of chlorine gas takes up 84 dm³ of volume (3.5 x 24).
- 🔄 The equation can be rearranged to calculate moles if the volume is known (volume ÷ 24 = moles).
- 💧 Example: 60 dm³ of oxygen equals 2.5 moles of oxygen (60 ÷ 24 = 2.5).
- 💡 For mass calculations, use the formula mass ÷ Mr (molar mass) to find moles, then use the gas equation to find volume.
- 🚰 Example: 27 grams of water vapor with an Mr of 18 equals 1.5 moles, which occupies 36 dm³ of volume (1.5 x 24).
- 🧫 When reacting gases, the volume of the product depends on the limiting reactant, in this case, nitrogen.
- 🔗 Use molar ratios from the balanced equation to calculate product volumes (e.g., 18 dm³ nitrogen forms 36 dm³ ammonia).
- ⚠️ The 24 constant only works for gases at room temperature and pressure; changes in temperature or pressure would affect the gas volume.
Q & A
What is the basic equation used to calculate the volume of a gas?
-The volume of a gas can be calculated by multiplying the number of moles of the gas by 24, as long as the gas is at room temperature and pressure. The result is the volume in decimeters cubed.
Does the type of gas affect the calculation of volume using this equation?
-No, the type of gas does not affect the calculation. Whether it’s chlorine, water vapor, or oxygen, the same equation can be applied.
How would you calculate the volume of chlorine gas if you have 3.5 moles of it?
-To find the volume of 3.5 moles of chlorine gas, you multiply the number of moles (3.5) by 24. The result is 84 decimeters cubed.
How can the equation be rearranged to find the number of moles from the volume of gas?
-To find the number of moles from the volume, the equation can be rearranged to 'volume ÷ 24 = moles.' For example, if you have 60 dm³ of oxygen, you divide 60 by 24 to get 2.5 moles of oxygen.
How do you find the volume of a gas if you're given its mass instead of moles?
-To find the volume from mass, first calculate the number of moles using the equation 'moles = mass ÷ relative formula mass (Mr).' Once you have the moles, multiply them by 24 to find the volume in decimeters cubed.
What is the process for finding the volume of water vapor if you are given 27 grams of it?
-First, calculate the moles of water vapor by dividing the mass (27 grams) by the relative formula mass of water (18). This gives 1.5 moles. Then multiply 1.5 moles by 24 to get a volume of 36 dm³.
How do you calculate the volume of ammonia produced from 18 dm³ of nitrogen in a reaction with excess hydrogen?
-First, calculate the moles of nitrogen by dividing 18 dm³ by 24, which gives 0.75 moles. Using the molar ratio of nitrogen to ammonia (1:2), you find that 0.75 moles of nitrogen will produce 1.5 moles of ammonia. Multiply 1.5 by 24 to get 36 dm³ of ammonia.
Can you simplify the process of calculating gas volumes when only gas volumes are involved?
-Yes, if you're only converting between gas volumes, you can use the molar ratio directly without finding moles. For instance, if the ratio of nitrogen to ammonia is 1:2, you can multiply the volume of nitrogen by 2 to get the volume of ammonia.
How much hydrogen would react with 4 dm³ of nitrogen based on the molar ratio?
-Since the molar ratio of nitrogen to hydrogen is 1:3, you multiply the nitrogen volume (4 dm³) by 3 to get 12 dm³ of hydrogen.
Does this equation with the number 24 work at any temperature and pressure?
-No, this equation with the number 24 only works for gases at room temperature and pressure. Changing the temperature or pressure would alter the gas volume.
Outlines
🔬 Understanding Gas Volume and Moles Relationship
In this section, the video explains a basic equation that links the volume of a gas (in decimeters cubed) to the number of moles. The formula is simple: multiply the number of moles by 24 to get the volume. This method applies to any gas, whether it’s chlorine, water vapor, or oxygen. For instance, 3.5 moles of chlorine gas take up 84 decimeters cubed. The reverse calculation is also possible by dividing the volume by 24 to find the number of moles. For example, 60 decimeters cubed of oxygen would equal 2.5 moles.
📏 Converting Mass to Volume Using Two Equations
The video continues by addressing more complex scenarios where moles aren't directly provided. To calculate the volume of gas from mass, another equation is introduced that links mass, moles, and the relative formula mass (Mr). The example used is 27 grams of water vapor. First, you find the number of moles by dividing the mass by the Mr of water (H2O), which is 18. This gives 1.5 moles. Finally, multiplying by 24, the volume is found to be 36 decimeters cubed.
🧪 Calculating Volume of a Product from a Reactant in a Reaction
This part of the video shifts focus to calculating the volume of a product gas (ammonia) when given the volume of a reactant gas (nitrogen). Given that nitrogen is the limiting reagent, the moles of nitrogen are calculated by dividing its volume (18 decimeters cubed) by 24. Using the molar ratio between nitrogen and ammonia (1:2), 0.75 moles of nitrogen would produce 1.5 moles of ammonia, and multiplying by 24 gives a final ammonia volume of 36 decimeters cubed.
🔄 Shortcut for Volume Calculations Using Molar Ratios
The video introduces a useful shortcut for gas volume conversions: instead of calculating moles, you can directly use the molar ratio between gases to find the volume. For instance, to find the volume of ammonia produced from 18 decimeters cubed of nitrogen, the molar ratio of 1:2 is used to multiply the nitrogen volume by 2, resulting in 36 decimeters cubed of ammonia. This shortcut works because doubling the moles of gas also doubles its volume.
⚖️ Using Molar Ratios for Other Gas Reactions
Another example is presented, where 4 decimeters cubed of nitrogen is used to calculate the volume of hydrogen it would react with. Using the molar ratio of nitrogen to hydrogen (1:3), you multiply the nitrogen volume by 3 to find that it would react with 12 decimeters cubed of hydrogen. This method only applies when converting between gas volumes, not moles or mass.
🌡 Limitations of the Gas Volume Equation at Different Conditions
The final point covered in the video highlights that the gas volume equation using the number 24 only works at room temperature and pressure. Changing either temperature or pressure would alter the volume a gas occupies. However, for exam purposes, this simplified equation using 24 is the only one needed. The video wraps up by encouraging viewers to practice with these principles for exam success.
Mindmap
Keywords
💡Volume
💡Moles
💡24
💡Room temperature and pressure
💡Relative formula mass (Mr)
💡Limiting reagent
💡Molar ratio
💡Gas volumes
💡Excess reactant
💡Mass
Highlights
Introduction of an equation that links the volume of a gas in decimeters cubed to the number of moles of gas.
The equation is simple because it works for any type of gas, regardless of its identity.
To calculate the volume of a gas, multiply the number of moles by 24 to get the volume in decimeters cubed.
Example given: 3.5 moles of chlorine gas would occupy 84 decimeters cubed.
Rearranging the equation allows for calculating moles when given a volume, such as 60 decimeters cubed of oxygen equating to 2.5 moles.
To find the volume of a gas given its mass, first calculate the moles using mass and relative formula mass.
Example of water vapor: 27 grams of water vapor with a relative formula mass of 18 gives 1.5 moles.
Using the original equation, 1.5 moles of water vapor occupies 36 decimeters cubed.
In a reaction where nitrogen reacts with excess hydrogen, the limiting reagent is nitrogen.
For 18 decimeters cubed of nitrogen, there are 0.75 moles.
Using the molar ratio of nitrogen to ammonia (1:2), 1.5 moles of ammonia is produced.
This results in 36 decimeters cubed of ammonia from the 18 decimeters cubed of nitrogen.
When only converting between gas volumes, it is unnecessary to calculate moles, as long as the molar ratio is known.
Example: 4 decimeters cubed of nitrogen reacts with 12 decimeters cubed of hydrogen, based on the molar ratio (1:3).
The equation with the number 24 only works for gases at room temperature and pressure.
Transcripts
in today's video we're going to be
focusing on this equation
which links the volume of a gas in
decimeters cubed
to how many moles of that gas we have
it's actually a really simple equation
because it doesn't matter which type of
gas you have
like whether you have chlorine or water
vapor or oxygen
all you have to do is multiply the
number of moles that you have
by the number 24
and that will give you the volume that
the gas takes up
measured in decimeters cubed
for example if we had 3.5 moles of
chlorine gas
then we would just do 3.5
times 24
to find that it took up 84 decimeters
cubed
we can also use the equation the other
way around
for example if we had 60 decimeters
cubed of oxygen
and we wanted to know how many moles
that was
we'd just rearrange the equation
to volume over 24 equals moles
and then do 60 divided by 24
which gives us 2.5 moles of oxygen
it can get a bit trickier if we're not
given the moles
for example if we had 27 grams of water
vapor
how would we find its volume
well we can see from our equation that
to find the volume we first need to find
the moles
so we're going to have to use this other
equation first
which links mass
moles and relative formula mass
already know that the mass is 27 grams
so we just need the mr
which for water which is h2o
would be 16 for the oxygen
plus 2 times 1 for the hydrogens
so eighteen
then we can divide the mass of 27 grams
by the mr of 18
to find that we must have 1.5 moles of
water vapor
then lastly all we need to do is go back
to our original equation
take our 1.5 moles
and multiply it by 24
to find that the water vapor would have
a volume of 36 decimeters cubed
another thing you might be asked to do
in the exam
is to find the volume of a product when
you're given the volume of a reactant
for example in this equation here
what volume of ammonia would be produced
if we reacted 18 decimeters cubed of
nitrogen with excess hydrogen
the first thing to notice is that the
hydrogen is in excess
which means that the nitrogen must be
the limiting reagent
and so the quantity of ammonia produced
is going to depend entirely on how much
nitrogen we have
next the normal thing to do in this sort
of question would be to find out how
many moles of nitrogen we have
so we take our volume of 18 decimeters
cubed
and divide it by 24
which tells us that we have
0.75 moles of nitrogen
and then we can use the molar ratio to
find out how many moles of ammonia that
will form
remember we find the molar ratio by
comparing these big numbers in front of
the chemical symbols
so an imaginary one for nitrogen because
when there's no number it just means
there's an unwritten one
and a two for ammonia
so the ratio is one to two
which tells us that for every mole of
nitrogen we have
we're going to make two moles of ammonia
so as we have 0.75 moles of nitrogen
we must have 0.75 times 2 moles of
ammonia
so 1.5 moles of ammonia
then finally we just go back to our
original equation
and multiply the 1.5 moles of ammonia by
24
to find that we'll make 36 decimeters
cubed of ammonia
now this whole idea of finding the moles
and then using molar ratios
is what you would normally have to do in
a question like this
however if you're only converting
between gas volumes like we are here
then you don't actually have to do any
of this stuff
all you need to do is look at the modal
ratio
which is one to two
and so you can multiply the original
volume of nitrogen which is 18
decimeters cubed
by two
to find that you'd make 36 decimeters
cubed of ammonia
and the reason we can do this is because
if we have twice as many moles of a gas
it's going to take up twice the volume
to see the zim practice let's change the
question a bit
if we start with four decimeters cubed
of nitrogen
how much hydrogen would it react with
because we're only converting gas
volumes we can just look at the molar
ratio between nitrogen and hydrogen
which is one two three
so then we just take the nitrogen's
volume of four decimeters cubed
and multiply it by 3
to find that it would react with 12
decimeters cubed of hydrogen
the very last thing i want to mention
is that this equation with the number 24
only works for gases at room temperature
and pressure
if we change the temperature or the
pressure
then it would change the number
as gases occupy different volumes at
different temperatures or pressures
but in the exams this equation with the
number 24 is the only one that you'll
need to use
anyway that's everything for this video
so hope you found it useful and we'll
see you again soon
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