Example of a “Level 3” type stoichiometry problem from
http://web.mst.edu/~gbert/reactor/Areactor.html
The starting screen looks like:
I clicked on the Level 3 button and then on “new reaction” and got:
The top bar graphs and mol values indicate the initial state of a system that contains
the indicated amounts of four substances, A, B, C, and D.
Click “run reaction” and you get:
Now you know what the final state of the system is AFTER reaction has occurred.
You are asked to determine the (standard) balanced chemical equation that represents
the reaction that occurred. This means you need to determine who the reactant and
product species are, as well as their coefficients such that the coefficients are in the
smallest whole number ratio.
How can you do this? Well, first note that:
1) the amounts of A and D have gone up,
2) the amount of B has gone down,
3) and the amount of C has not changed
This means that:
1) A and D are being produced as reaction occurs.  They are products
2) B is being consumed (“used up”) as reaction occurs.  It is a reactant
3) C is neither a reactant nor a product (no net change. It won’t appear in the
balanced equation representing this reaction)
To get the coefficients, you need to first calculate how many moles of each species
has changed. Then you need to look at the ratio of those changes. One way to help
“see” all of this in a compact and organized way is to use an ICF table (“I” stands for
initial, “C” stands for change, and “F” stands for final). For the current example, the
table would initially look like this (based on the values from the applet):
(values in moles)
A
B
C
D
I (initial)
4.1
3.2
8.4
1.3
9.7
0.4
8.4
9.7
C (change in)
F (final)
As you may have learned in a prior math or science course, the change in a variable,
often expressed as a , is defined as “final” – “initial”.
So A here (in moles) = 9.7 mol – 4.1 mol = +5.6 moles.
Similarly, B is 0.4 – 3.2 = -2.8 mol (NOTE: the negative sign here means that the
amount of B decreases as time goes by),
C is 8.4 – 8.4 = 0 mol (its value did not change, right?)
D is 9.7 – 1.3 = +8.4 mol
The table now becomes:
(values in moles)
A
B
C
D
I (initial)
4.1
3.2
8.4
1.3
C (change in)
+5.6
-2.8
0
+8.4
F (final)
9.7
0.4
8.4
9.7
[Note that the signs are consistent with our earlier conclusion that A and D form and
thus are products, and B decreases and is thus a reactant.]
** It is the ratio of the “changes” that equals the ratio of the coefficients in the
balanced chemical equation, because that’s what the coefficients in a
balanced equation represent—the ratio in which reactants react and products
form (in moles)!
So, to reduce this ratio to a whole-number one, you can either do it by “inspection”, or
do what you did when you first learned how to get an empirical formula from “moles of
atoms” values—divide each value by the smallest (not including the zero, obviously).
So here, divide +5.6, -2.8, and 8.4 by 2.8, and you’ll get 2, -1, and 3, so the coefficients
of A, B, and D respectively are 2, 1, and 3. Being careful to recall that B is the reactant
and A and D products, the balanced equation follows:
B → 2A+3D
If you input that into the program (see below), you’ll see that it says “RIGHT” in colored
letters to indicate you have entered the correct answer:
Just for completeness, if you input an incorrect answer, you will get a “sorry” response,
as well as a “tutor” button, which if clicked, will simply give you the correct answer
(unfortunately, no “tutoring”).