UNIT 10 REVIEW WORKSHEET
1.
Convert the following pressure measurements to atmospheres.
(A)  151.98 kPa           (B)  456 mm Hg            (C)  912 torr
(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)