Chemistry Reference
In-Depth Information
volume
molar volume
33.5 d
m
3
22.4 dm
3
/mole
= 1.50 moles
19. [48.0 g]—
# of moles =
=
32.0 g
mole
mass = # of moles × molar mass = 1.50 moles ×
= 48.0 g
volume
molar volume
10.0 d
m
3
22.4 dm
3
/mole
= 0.446 moles
20. [2.68 × 10
23
]—
# of moles =
=
6.02 × 10
23
molecule
mole
# of molecules = 0.446 moles ×
= 2.68 × 10
23
mol.
21. [98.8 dm
3
]—
C
3
H
8
+ 5O
2
3CO
2
+ 4H
2
O
38.8 g
x dm
3
mass
molar mass
38.8 g
44.0 g/mole
# of moles C
3
H
8
=
=
= 0.882 moles
A.
B. # of moles of O
2
can be determined from the molar ratio:
C
3
H
8
O
2
coefficients
1
0.882
5
x
# of moles
=
# of moles of O
2
= 4.41 moles
22.4 dm
3
mole
C.
22. [33.3 dm
3
]—
Volume of O
2
= # of moles × molar volume = 4.41 moles ×
= 98.8 dm
3
3CO
2
+ 4H
2
O
x dm
3
20.0 dm
3
The volume of O
2
can be determined from the molar ratio:
C
3
H
8
+ 5O
2
O
2
CO
2
coefficients
5
x
3
20.0 dm
3
=
volume
volume of O
2
= 33.3 dm
3
23. [66.1 g]—
C
3
H
8
+ 5O
2
3CO
2
+ 4H
2
O
40.0 g
x g
mass
molar mass
40.0 g
44.0 g/mole
A.
# of moles C
3
H
8
=
=
= 0.909 moles
B. # of moles of H
2
O can be determined from the molar ratio:
C
3
H
8
H
2
O
coefficients
# of moles
1
0.909
4
x
=
# of moles of H
2
O = 3.67 moles
18.0 g
mole
C.
24. [43.1 g]—
Mass of H
2
O = # of moles × molar mass = 3.67 moles ×
= 66.1 g
2MgO
30.0 g 12.0 dm
3
2Mg + O
2
x g
mass
molar mass
30.0 g
24.3 g/mole
A.
# of moles of Mg =
=
= 1.23 moles
volume
molar volume
12.0 d
m
3
22.4 dm
3
/mole
= 0.536 moles (limiting)
B.
# of moles of O
2
=
=
C. The oxygen is the limiting reactant because there is more than twice as much
magnesium than oxygen.
D. # of moles of MgO can be determined from the molar ratio:
O
2
MgO
coefficients
1
0.536
2
x
=
# of moles
# of moles of MgO = 1.07 moles
40.3 g
mole
= 43.1 g
E. Mass of MgO = # of moles × molar mass = 1.07 moles ×
