Chemistry Reference
In-Depth Information
Example 2
What volume of oxygen gas at STP will be generated from the
decomposition of 35.0 g of water?
Again, you have to come up with the balanced chemical reaction. Based
on what you should know about decomposition reactions (see Lesson 6-2)
and balancing equations (see Lesson 6-1), this shouldn't be too much
trouble. Complete Step 1, starting with a balanced equation, and Step 2,
labeling the equation.
What volume of oxygen gas at STP will be generated from the decom-
position of 35.0 g of water?
2H
2
O
2H
2
+ O
2
35.0 g x dm
3
Notice that this time the unknown is labeled with the units dm
3
. We
don't use the unit “g” this time, because the problem makes it quite clear
that a volume is being sought. Of the various units of volume, dm
3
is the
natural choice, because this is the unit that we have been using with the
molar volume of gases.
For Step 3, you must convert the given quantity into the number of
moles. Because the given quantity is in grams, you'll use the molar mass of
water to convert to moles. Once again, find the molar mass of water (18.0
g/mole) on the periodic table.
2H
2
O
2H
2
+ O
2
35.0 g
dm
3
mass of sample
molar mass
35.0 g
18.0 g/mole
= 1.94 moles
# of moles of H
2
O =
=
Next, make use of the molar ratio of water to oxygen to determine the
number of moles of oxygen we would produce. The coefficients show that
the ratio is 2:1, which means that it takes 2 moles of water to produce 1
mole of oxygen. It follows that 1.94 moles of water will produce 0.970 moles
of oxygen. If you don't see that, look at the ratio here:
2H
2
OO
2
coefficients
2
1.94
1
x
=
# of moles
We cross-multiply to find that 2x=1.94, which means that x (the num-
ber of moles of O
2
) = 0.970 moles.
Finally, multiply the number of moles of oxygen by the volume of each
mole of oxygen at STP, in dm
3
.
