Chemistry Reference
In-Depth Information
B. # of moles of O
2
can be determined from the molar ratio:
Mg
O
coefficients
# of moles
2
0.144
=
1
x
# of moles of O
2
= 0.0720 moles
C. Volume of O
2
= # of moles × molar volume = 0.0720 moles × 22.4 dm
3
/mole
= 1.61 dm
3
8.
[2.52 g]—
2C
2
H
6
+ 7O
2
4CO
2
+ 6H
2
O
1.00 dm
3
x
g
volume
molar volume
1.00 d
m
3
22.4 dm
3
/mole
= 0.0446 moles
A.
# of moles of C
2
H
6
=
=
B. # of moles of H
2
O
can be determined from the molar ratio:
C
2
H
6
H
2
O
coefficients
2
0.0466
=
6
x
# of moles
# of moles of H
2
O
= 0.140 moles
C. Mass of H
2
O = # of moles × molar mass = 0.140 moles × 18.0 g/mole = 2.52 g
Lesson 7-6 Review
1.
[25 dm
3
of oxygen]—
C
3
H
8
+ 5O
2
4H
2
O + 3CO
2
5.0 dm
3
x dm
3
propane
oxygen
coefficients
1
5.0 dm
3
=
5
x
volume
So, the volume of the oxygen (x) = 25 dm
3
2.
[9.0 dm
3
of carbon dioxide]— C
3
H
8
+ 5O
2
4H
2
O + 3CO
2
15 dm
3
x dm
3
O
2
CO
2
coefficients
volume
5
15 dm
3
=
3
x
So, the volume of the CO
2
(x) = 9.0 dm
3
3.
[120. dm
3
of water]—C
3
H
8
+ 5O
2
4H
2
O + 3CO
2
x dm
3
90.0 dm
3
H
2
O
2
coefficients
4
x
3
90.0 dm
3
=
volume
So, the volume of the H
2
O (x) = 120. dm
3
4.
[6.0 dm
3
of nitrogen]—
3H
2
+ N
2
2NH
3
x dm
3
4.0 dm
3
H
2
N
2
coefficients
volume
3
x
2
4.0 dm
3
=
So, the volume of the H
2
O (x) = 6.0 dm
3
5.
[5.0 dm
3
of oxygen]—
C
6
H
12
O
6
+ 6O
2
6H
2
O + 6CO
2
x dm
3
5.0 dm
3
O
2
CO
2
coefficients
6
x
6
5.0 dm
3
=
volume
So, the volume of the O
2
(x) = 5.0 dm
3
