RICE TABLES
PURPOSE OF
RICE TABLES
• While RICE tables can support basic
stoichiometry, It is especially helpful to
support two functions:
– Identification of Limiting Reagents
(reactants)
– Shifts in equilibrium
• NOTE: The calculations for equilibrium
are beyond the scope of this course so
we will only focus only on its use for
identifying limiting reagents.
RICE TABLE
• A way to organize information
regarding the moles of substances in a
chemical reaction.
• Pneumonic device to recall its contents:
– R = Reaction Ratios
– I = Initial conditions
– C = Changes during the reaction
– E = Endpoint conditions of the
reaction
Once the chemical equation is balanced, the coefficients
establish the mole ratios for the reaction.
+
R
I
C
E
→
+
2 K3PO4
3 CaCl2
6 KCl
Ca3(PO4)2
2 mol
3 mol
6 mol
1 mol
The initial conditions identify the number of moles of each
substance available for the reaction. Before the chemical
reaction has started, no products can exist.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
0 mol
0 mol
To identify the limiting reactant, you will have information for each of the
reactants. If you are not provided that information in moles, you will need
to convert to moles first.
For the purposes of this example, we will use 2.4 moles of potassium
phosphate and 2.7 moles of calcium chloride as the initial conditions.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
Unless you can readily identify the limiting reactant, assume that the
reactant listed first is the one that is limited. If true, all of that reactant will
be completely used.
Therefore, the reaction will consume all of that reactant. This is signified
by subtracting all of that reactant.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-2.4 mol
Now you will use the mole ratio to determine how many moles of the other
reactant will be consumed based on the use of all of the 1st reactant.
In this case the mole ratio is 2:3. Therefore 2.4 moles of the first reactant
will consume 3.6 moles of the other reactant. Enter this value in the RICE
table, and as it is consumed, it will be subtracted.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-2.4 mol
-3.6 mol
The ending amount will be the result of the application of the
mathematical operation of the initial condition and the change. Since the
change removes reactant, it will be a subtraction.
Notice that more calcium chloride was used than was available. Since this
is impossible, the wrong reactant was selected as the limiting reactant.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-2.4 mol
-3.6 mol
0
-0.9 mol
X
Since the wrong reactant was initially selected, you will need to start over
using the other reactant, repeating the entire process.
Using calcium chloride as the limiting reactant you would get the following
after re-doing all the steps. Not only have you found the limiting reactant,
you have also identified how much potassium phosphate remains.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-1.8 mol
-2.7 mol
0.6 mol
0 mol
Now that the limiting reactant is identified, it is possible to determine the
amount of products produced by using the mole ratios for potassium
chloride and calcium phosphate.
Since products are produced they are added to the system.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-1.8 mol
-2.7 mol
+5.4 mol
+0.9 mol
0.6 mol
0 mol
Since there were no products prior to the start of the reaction, the added
values are the theoretical values for the products of the reaction.
If the calculation asks for mass instead of moles, convert the moles to mass.
2 K3PO4 + 3 CaCl2 → 6 KCl + Ca3(PO4)2
R
I
C
E
2 mol
3 mol
6 mol
1 mol
2.4 mol
2.7 mol
0 mol
0 mol
-1.8 mol
-2.7 mol
+5.4 mol
+0.9 mol
0.6 mol
0 mol
5.4 mol
0.9 mol
Theoretical yields