Chemistry Reference
In-Depth Information
Lesson 7-5: Mixed Mass-Volume Problems
In problems involving a gas, it is easier to measure the volume of the gas
than the mass. Mixed mass-volume problems include both mass-volume prob-
lems, where you are given a mass and asked for a volume, and volume-
mass problems, where you are given a volume and asked for a mass. You
will need to do the molar conversions involving the molar volume of a gas,
so review Lesson 7-2 if necessary. The basic steps for solving this type of
problem remain the same:
1.
Begin with a balanced chemical reaction.
2.
Carefully label the equation with the information from the
word problem.
3.
Convert the given quantity to number of moles.
4.
Use the molar ratio shown by the coefficients to determine
the number of moles of the unknown.
5.
Convert this number of moles to the quantity the question
asks for.
Let's try an example.
Example 1
How many grams of zinc must be completely reacted with an excess
of sulfuric acid according to the balanced reaction shown here, in
order to generate 500 cm
3
of hydrogen gas, at STP?
Zn + H
2
SO
4
ZnSO
4
+H
2
You have a balanced chemical reaction, which takes care of Step 1, and
the fact that all of the coefficients are 1 means that the step with the molar
ratio will be easy. There is one potential tricky part to this problem, however.
Can you see what it is? The volume of the gas that they want to generate is
reported in cm
3
. The molar volume of a gas is 22.4 dm
3
/mole. You must
convert 500 cm
3
to dm
3
before you can convert it to the number of moles.
You should remember that there are 1000 cm
3
in one dm
3
, so you can
make the conversion with the factor-label method, as shown here:
1 dm
3
1000 cm
3
= 0.5 dm
3
500 cm
3
×
Now, label the problem, which is the second step in this type of prob-
lem. Notice how I cross out the original volume, and replace it with its
equivalent in dm
3
.
