Chemistry Reference
In-Depth Information
The first task in doing a problem involving a limiting quantity is to recog-
nize that it is such a problem. Fortunately, that is fairly easy: The quantities of
two different reactants are given. Then, use the tabulation method outlined in
Section 10.1. Do the steps given in the left column to solve the following exam-
ple in the right column:
Calculate the number of moles of each of the products and of the excess
reactant when 1.500 mol of A and 0.500 mol of B are allowed to react accord-
ing to the following general equation:
Limiting quantities problems
have the quantities of two (or
more) reactants given.
3 A
2 B £ C
3 D
Steps
Example
Step 1:
Write the balanced chemical equation for the
reaction.
3 A
2 B
£
C
3 D
Step 2:
Write the initial number of moles of each reac-
tant (and product) under its formula.
All quantities are in moles.
Present initially
1.500
0.500
0
0
Step 3:
Determine which reactant is limiting by divid-
ing the number of moles of each reactant given
in the problem by the coefficient of that reactant
in the balanced chemical equation. The smallest
quotient indicates the limiting reagent. Then
draw a line through the results so as not to use
them later. Rewrite the number of moles pres-
ent of the limiting quantity with a minus sign
in a row corresponding to the change due to
reaction.
1.500 mol A
3 mol A
0.500 mol B
2 mol B
0.500
0.250
B is limiting.
Change due
to reaction
0.500
Step 4:
Complete the “change” row by writing the
number of moles of each substance that would
react with or be produced from the quantity in
step 3. Use a minus sign with each quantity of
reactant. The magnitudes in the “change” row
are in the same ratio as the coefficients in the
balanced chemical equation.
Change due
to reaction
0.750
0.500
0.250
0.750
Step 5:
Subtract the quantity of each reactant in the
“change” row from the initial quantity and add
the quantity of each product in the change row
to any initial quantity to get a final quantity.
Present finally
0.750
0.00
0.250
0.750
Using this method, we calculate the quantity of each product produced and the
quantity of any excess reactant all in one calculation. Note that the quantities
in the “initial” and “final” rows are
not
in the ratio of the balanced chemical
equation; only the magnitudes of the ratios in the “change” row are in the same
ratio as those in the balanced equation.
