Chemistry Reference
In-Depth Information
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
mass of sample
molar mass
35.0 g
18.0 g/mole
= 1.94 moles
# of moles of H
2
O =
Volume of O
2
= # of moles × molar volume of gas at STP
= 0.970 moles × 22.4 dm
3
/mole = 21.7 dm
3
=
You find that 35.0 g of water will decompose to form 21.7 dm
3
of oxy-
gen, at STP. Before you go on to the practice problems, let's add some-
thing to the previous problem. How many dm
3
of hydrogen at STP would
be produced from the decomposition of 35.0 g of water? Can you quickly
see why the answer must be 43.4 dm
3
? You see, the coefficient of “2” in
front of the hydrogen tells you that you are going to produce twice as many
moles of hydrogen as oxygen. Or, another way to look at it, you are going
to get as many moles of hydrogen out as you put of water in.
S
o, you will
generate 1.
9
4 moles of hydrogen gas, which represents 43.4 dm
3
(1.94 moles
× 22.4 dm
3
/mole). The fact that doubling the volume of oxygen (2 × 21.7dm
3
= 43.4 dm
3
) gives us this same answer is a clue to a shortcut that I will teach
you in the next lesson.) Answer these practice problems before moving on
to the next lesson.
Lesson 7-5 Review
Use the following balanced chemical equation to answer questions 1-3:
3H
2(g)
+ N
2(g)
2NH
3(g)
hydrogen + nitrogen
ammonia
1.
How many dm
3
of hydrogen gas at STP would be required to produce
50.0 g of ammonia?
2.
How many grams of ammonia can be produced when 23.5 dm
3
of
nitrogen gas at STP react with an excess of hydrogen?
3.
How many dm
3
of hydrogen gas at STP will react with exactly 56.0
grams of nitrogen gas?
Use the following balanced chemical equation to answer questions 4-6:
2H
2
O
(l)
2H
2(g)
+ O
2(g)
water
hydrogen + oxygen
4.
How many grams of water would be required to produce 30.0 dm
3
of
oxygen gas at STP?
