As I stated in the lesson 9-3,
the coefficients in chemical equations provide us with some important information.
They show us the molar ratio relationships that exist between the reactants and
products. It is very important for you to remember that the coefficient show the
ratio of particles-particles and of moles-moles, not mass-mass. For example,
look at the reaction shown below;
2H2(g) + O2(g)
----> 2H2O(g)
The coefficients tell us that 2
molecules of hydrogen react with 1 molecule of oxygen to form 2 molecules of water.
They also tell us that 2 moles of hydrogen will react with 1 mole of oxygen to form 2
moles of water. These coefficients do not tell us the ratio of
mass-mass, meaning 2 grams of hydrogen do not react with one gram of
oxygen to form 2 grams of water! That would not make sense!
However,
when dealing with both a gaseous given and a gaseous unknown, we can take advantage of a
special property of gases to make our calculations much simpler. Avogadro's law
tells us that "at equal temperatures and pressures, equal volumes of different gases
contain the same number of particles." This lead to the idea of a standard for
the volume of one mole of gas. By now, probably know that one mole of any gas at
standard temperature and pressure has a volume of 22.4 dm3. This
constant, 22.4 dm3, is called the molar volume of a gas. Each element or
compound has its own individual molar mass, but the molar volume of all gases is the
same. This makes stoichiometry problems involving only gases much easier to
do. To demonstrate, I will show you an example of solving a volume-volume problem
the long way, and then the same problem with the shortcut.
To solve Volume-Volume problems
the long way, we would use a modified version of the three steps from last lesson:
1. Change the volume of the given gas to number of
moles by using the formula;
volume of gas at STP
# of moles = -----------------------------------------------
molar volume of gases (22.4 dm3/mole)
2. Determine the number of moles of
the unknown by comparing the molar ratio;
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
3. Change the number of moles of the
unknown gas to volume at STP by using the formula;
volume of a gas at STP = # of moles of gas
x molar volume of gas (22.4 dm3/mole)
Now, if you are a
good Math student you can probably see why we can take a shortcut. Steps 1 and 3
call for opposite operations (x and / by 22.4) so they will cancel each other out!
For the sake of the students who don't see that right away, I will demonstrate the Math
that you would end up doing if you followed the three steps above.
Example 1. What volume of oxygen gas
would react with 35.0 dm3 of hydrogen gas at STP, according to the equation
below?
2H2(g) + O2(g)
----> 2H2O(g)
First, label the given and the unknown, and
cross out the other information, which is only useful when balancing the equation;
given unknown
2H2(g) + O2(g) ----> 2H2O(g)
35.0 dm3 Xdm3
Now, we will follow the three steps shown
above. Remember, this will be the long way to do things. I am only doing this
for demonstration purposes!
1. Change the volume of the given gas to number of
moles by using the formula;
Volume of given gas (hydrogen) at STP = 35.0 dm3
Molar volume of gases = 22.4 dm3
volume of gas at STP
# of moles = -----------------------------------------------
molar volume of gases (22.4 dm3/mole)
35.0 dm3
# of moles = ------------
22.4 dm3/mole
# of moles of hydrogen gas = 1.56 moles
2. Determine the number of moles of
the unknown by comparing the molar ratio (from balanced equation);
Number of moles of given (hydrogen) = 1.56
moles
Coefficient of given (hydrogen) = 2
Coefficient of unknown (oxygen) = 1
number of moles of given
number of moles of unknown
-------------------------------- =
--------------------------------------
coefficient of given coefficient of unknown
1.56 moles
X moles
------------ = ------------
2
1
2x = 1.56 moles
# of moles of the unknown (oxygen) = 0.780
moles
3. Change the number of moles of the
unknown gas to volume at STP by using the formula;
# of moles of oxygen (unknown) = 0.780 moles
Molar volume of a gas at STP = 22.4 dm3/mole
volume of a gas at STP = # of moles of gas
x molar volume of gas (22.4 dm3/mole)
Volume of the oxygen gas at STP = 0.780 moles
x 22.4 dm3/mole
Volume of the oxygen gas at STP = 17.5 dm3
So, our final answer is that 17.5 dm3
of oxygen gas would react with 35.0 dm3 of hydrogen gas, at STP, according to
the reaction;
2H2(g) + O2(g)
----> 2H2O(g)
Do you notice that the ratio of volume-volume is identical to the ratio of
moles-moles! This is because the opposite operations that I mentioned earlier.
In step one, we divided by 22.4. In step three, we multiplied by 22.4.
These steps cancel each other out, so they do not need to be shown! When you are
working with only gases, at equal temperature and pressure, all you have to do is compare
the ratios shown in a balanced chemical equation. In other words, the coefficient in
a chemical equation show the ratio of particles-particles, moles-moles, and volume-volume
(when dealing with gases at the same temperature and pressure.) To solve
volume-volume problems, use the formula below;
volume of the given
volume of the unknown
---------------------------------- =
----------------------------------
coefficient of the given coefficient of the unknown
Let us go back and solve the original
problem the easy way:
Example 2. What volume of oxygen gas
would react with 35.0 dm3 of hydrogen gas at STP, according to the equation
below?
2H2(g) + O2(g)
----> 2H2O(g)
First, identify the given and the unknown,
and cross out the extra information;
given unknown
2H2(g) + O2(g) ----> 2H2O(g)
35.0 dm3 Xdm3
volume of the given (hydrogen) = 35.0 dm3
volume of the unknown (oxygen) = X dm3
coefficient of the given (hydrogen) = 2
coefficient of the unknown = 1
volume of the given
volume of the unknown
---------------------------------- =
----------------------------------
coefficient of the given coefficient of the unknown
35.0 dm3
Xdm3
---------- = ---------
2 1
2X = 35.0 dm3
(X) volume of the oxygen = 17.5 dm3
So, we get the same
answer with much less work! Imagine how much time this will save you on tests where other
students are solving volume-volume problems the long way! Please remember -
***The above shortcut only works
with volume-volume problems!***
Now, practice
what you have learned with the worksheets and links below.
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
