Chemistry Reference
In-Depth Information
Practice Problem 10.4
In a certain reaction, 0.225 mol of
H
2
gas
reacts partially with
N
2
gas to yield gaseous
NH
3
.
If 0.033 mol of
H
2
remains
after the reaction is stopped, how many moles of
N
2
is used up?
Tabulation Method
Many times (Section 10.4, Chapters 18 and 19) we will find it useful to tabu-
late the numbers of moles (or related quantities to be introduced later) of the
substances undergoing reaction. For example, the data of Example 10.2 can be
tabulated as follows:
4 Al(s)
3 O
2
(g)
£
2 Al
2
O
3
(s)
Quantities must be in moles.
Initial quantities
Excess
3.18 mol
0.00 mol
Change due to reaction
4.24 mol
3.18 mol
2.12 mol
It must be noted that the magnitudes of the quantities in the “change due to reac-
tion” line are
always
in the ratio of the coefficients in the balanced chemical
equation. It will also become apparent that the numbers of moles of reactants in
the “change” line are
subtracted
from the initial quantities present and the num-
bers of moles of products are
added
to any initial quantities present.
4 Al(s)
3 O
2
(g)
£
2 Al
2
O
3
(s)
Initial quantities
Excess
3.18 mol
0.00 mol
Change due to reaction
Final quantities
4.24 mol
3.18 mol
2.12 mol
Some excess
0.00 mol
2.12 mol
EXAMPLE 10.5
Calculate the quantities of the salt and water in solution after the reaction of
1.33 mol of HCl and excess NaOH in 15.17 mol of water.
Solution
HCl(aq)
NaOH(aq)
£
H
2
O(
)
NaCl(aq)
Initial quantities
1.33 mol
Excess
15.17 mol
0.00 mol
Change due to
reaction
Final quantities
1.33 mol
1.33 mol
1.33 mol
1.33 mol
0.00 mol
Some excess
16.50 mol
1.33 mol
Snapshot Review
❒
The balanced chemical equation gives the mole ratios of all the
substances in the reaction, just as an empirical formula gives the
ratios of atoms of the elements in a compound. As with chemical
formulas, these ratios can be used as factors in calculations involving
any two of the substances.
ChemSkill Builder
4.1, 4.3
