UNIT 10 REVIEW
WORKSHEET
1.
Convert the following pressure measurements to atmospheres.
(A) 151.98 kPa
| 1 atm
= 1.5003 atm
| 101.3 kPa
(B) 456 mm Hg | 1 atm =
0.600 atm
| 760 mm Hg
(C) 912 torr
| 1 atm =
1.20 atm
| 760 torr
2.
What are the standard conditions for gas measurement?
Standard Temperature
& Pressure (STP) is 0 oC & 1 atm
3. The
volume of a sample of methane gas measures 350. mL at
27.0 oC and 810. mm
Hg.
What is the volume (in liters)
at -3.0 oC and 650.
mm Hg pressure?
P1 V1 = P2
V2 (810 mm Hg)
(350 mL) = (650 mm Hg) (x) x is V2
T1 T2 300 K
270 K
cross multiplying… 195000
x = 76545000
solving
for x… x =
393 mL
4.
How many grams of nitrogen gas are contained in a 32.6 liter container at 34.4 oC
and 579 torr?
PV = nRT P = pressure in atm = 579 torr
| 1 atm =
0.762 atm
| 760 torr
V
= volume in liters = 32.6 L
n = quantity in moles = ?
R = ideal gas constant = 0.0821
L atm/mole K
T = temperature in Kelvin = 34.4
+ 273 = 307.4 K
solving for n… n = P V
R T
substituting… n = 0.762 .
32.6 = 0.984 moles
0.0821 . 307.4
converting to grams… 0.984 moles .
28 g/mole = 27.6 g
MM of nitrogen gas (N2)… N: 2 x 14 = 28
5. A mixture of
four gases in a container exerts a total pressure of 955 mm Hg. In
this
container, there are 4.50 moles of nitrogen gas, 4.25 moles of carbon
dioxide gas, 2.75
moles of hydrogen gas, and 2.00 moles of oxygen gas. What is the
partial pressure of
each gas?
Px = (moles x) . Ptotal
(total moles)
total moles = 4.50 + 4.25 + 2.75 + 2.00
= 13.50 moles
N2 = 4.50 moles .
955 = 318 mm Hg CO2
= 4.25 moles . 955 = 301 mm Hg
13.50 moles 13.50
moles
H2 = 2.75 moles .
955 = 195 mm Hg O2
= 2.0 moles . 955 = 141 mm Hg
13.50 moles 13.50 moles
6. Compare the rates of effusion of carbon
dioxide gas and carbon monoxide gas.
rate A = A
= lighter gas = CO
rate B B = heavier gas = CO2
= 1.25
So, CO (lighter gas) effuses 1.25 times
faster than CO2 (heavier gas).
7. An unknown gas effuses 1.37 times faster
than chlorine gas. What is the molar
mass
of the unknown gas?
rate A =
rate B
A
= lighter gas = unknown (You know it is
lighter because it effuses faster.)
B = heavier gas = chlorine
(Cl2)
Substituting… 1.37 =
1
Solve by first squaring both sides… 1.88 = 71
1
x
Cross multiply… 1.88 x = 71
x = 37.8 g/mole (“g/mole” is the unit for molar mass)
8. Given the following unbalanced reaction:
___ C5H12 +
___ O2 à ___ CO2 + ___
H2O
How many liters of oxygen are needed to
produce 45.7 liters of CO2?
___ C5H12 +
_8_ O2 à _5_ CO2 + _6_
H2O
x L 45.7 L
45.7 L CO2 = x L O2
5 8 x
= 73.1 L O2
9. Given the unbalanced equation:
___ Mg +
___ O2 à ___ MgO
How many liters of oxygen gas are
required to produce 45.8 grams of magnesium oxide?
_2_ Mg +
___ O2 à _2_ MgO
x
L 45.8 g
45.8 g MgO
| 1 mole MgO = 1.136 moles MgO 1.136 moles MgO
= x moles O2
|40.3 g MgO 2 1
X = 0.568
moles O2| 22.4 L O2 = 12.8 L O2
|1 mole O2
10. An
aerosol can contains gases under a pressure of 4.50 atm at 20.0 oC. If the can
is left on a hot, sandy beach,
the pressure of the gases increases to 4.80 atm.
What is the temperature on the
beach (in oC)?
P1 = P2 4.50
= 4.80 Cross
multiply… 4.50 x = 1406.4
T1 T2 293 x Solve
for x… x = 313 K
Convert
to Celsius… 313 – 273 = 40 oC
11.) D = P . MM D = (1 atm) (36.5 g/mole)
D = 1.63 g/L
R .
T (0.0821 L.atm/mole.K)
(273 K)
12.) 30.17 in
Hg | 2.54 cm = 76.6318 cm
Hg x
10 mm = 766.3 mm Hg (atmospheric)
| 1 inch 1 cm
766.3 mm Hg –
19.2 mm Hg = 747.1 mm Hg 747.1
mm Hg | 1 atm = 0.9830 atm
(atmosphere) (water) | 760 mm Hg
21.5 oC + 273 = 294.5 K 250. mL = 0.250 L
MM = gRT (1.27
g) (0.0821 L.atm/mole.K) (294.5 K) = 125 g/mole
PV (0.9830 atm) (0.250
L)