Download GAS LAW PROBLEMS

Honors Chemistry
Name ____________________________
Chapter 13: Gas Laws and Stoichiometry WS II
Date _____/_____/_____
Period _____
1. Calcium hydride reacts with water to form hydrogen gas and calcium hydroxide. This reaction is
sometimes used to inflate life rafts, weather balloons, and the like where a simple, compact means of
generating H2 is desired. How many grams of calcium hydride are needed to generate 10.0 L of H2
gas if the partial pressure of H2 is 740 mm Hg at 23.0 degrees C?
CaH2
+
2 H2O

Ca(OH)2
+
2 H2
n = PV = (0.97 atm) (10.0 L) (mole ▪ K) = 0.40 moles H2
RT
0.40 moles H2
(0.0821 L ▪ atm) (296 K)
1 moles CaH2
2 moles H2
42.10 g CaH2
1 moles CaH2
= 8.4 g CaH2
2. The metabolic breakdown of glucose, C6H12O6, in our bodies produces CO2, which is expelled from
our lungs as gas: C6H12O6 (s) + O2 (g)  CO2 (g) + H2O () (equation is unbalanced) Calculate the
volume of dry CO2 gas produced at body temperature (37.0 degrees C) and 1.00 atm when 5.00
grams of glucose is consumed in the reaction.
C6H12O6 + 6 O2  6 CO2 + 6 H2O
5.00 g C6H12O6
1 moles C6H12O6
180.2 g C6H12O6
6 moles CO2
1 moles C6H12O6
= 0.166 moles CO2
V = nRT = (0.166 mole) (0.0821 L ▪ atm) (310. K) = 4.5 L CO2
P
(mole ▪ K) (0.94 atm)
3. Ammonium sulfate, an important fertilizer, can be prepared by the reaction of ammonia with sulfuric
acid. Calculate the volume of ammonia needed at 20. degrees C and 25.0 atm to react with 150 kg
of sulfuric acid.
2 NH3 + H2SO4  NH4SO4
150 Kg H2SO4
1000 g H2SO4
1 kgH2SO4
1 mole H2SO4
98.09 g H2SO4
2 mole NH3
1 mole H2SO4
V = nRT = (3100 mole) (0.0821 L ▪ atm) (293 K) = 3.0 x 103 L NH3
P
(mole ▪ K) (25.0 atm)
= 3100 moles NH3
4. Hydrogen gas and zinc sulfate is produced when zinc reacts with sulfuric acid. If 124 mL of wet H2
gas is collected over water at 24 degrees C and a barometric pressure of 725 mm Hg, how many
grams of Zn have been consumed? (the vapor pressure of water at 24 degrees C is 22.38 mm Hg).
Zn + H2SO4  H2 +
ZnSO4
725 mm – 22.38 mm = 703 mm Hg = 0.925 atm
n = PV = (0.925 atm) (0.124 L) (mole ▪ K) = 0.00470 moles H2
RT
(0.0821 L ▪ atm) (297 K)
0.00470 moles H2
1 mole Zn
1 mole H2
65.39 g Zn
1 moles Zn
= 0.307 g Zn
5. In the reaction, N2 + H2  NH3, how many mL of nitrogen, measured at STP, are required to produce
400 mL of NH3, measured at STP? How many mL of hydrogen are required at STP?
N2
+
3 H2

2 NH3
0.4 L NH3
1 mole NH3
22.4 L NH3
1 mole N2
2 mole NH3
22.4 L N2
1 mole N2
1000 mL N2
1 L N2
= 200 mL N2
0.4 L NH3
1 mole NH3
22.4 L NH3
3 mole H2
2 mole NH3
22.4 L H2
1 mole H2
1000 mL H2
1 L H2
= 600 mL H2
6. In the reaction, NO + H2  H2O + N2, how many mL of nitrogen, measured at STP, would be
produced from 0.00013 grams of hydrogen?
2 NO
0.00013 g H2
1 mole H2
2.02 g H2
+
2 H2

2 H2O
1 mole N2
2 mole H2
+
N2
22.4 L N2
1 mole N2
1000 mL N2
1 L N2
= 0.72 mL N2
7. Calculate the maximum number of mL of CO2, at 740 torr and 28 degrees C, that could be produced
by reacting 500 mL of CO (at 760 torr and 15 oC) with 500 mL of O2 (at 770 torr and 0 oC).
2 CO
+
O2

2 CO2
n = PV = (1.00 atm) (0.5 L) (mole ▪ K) = 0.02 moles CO
RT
(0.0821 L ▪ atm) (288 K)
n = PV = (1.01 atm) (0.5 L) (mole ▪ K) = 0.02 moles O2
RT
(0.0821 L ▪ atm) (273 K)
0.02 moles CO
2 moles CO2
2 moles CO
0.02 moles CO2
0.02 moles O2
2 moles CO2
1 moles O2
= 0.04 moles CO2
V = nRT = (0.02 mole) (0.0821 L ▪ atm) (301 K) = 0.5 L = 500 mL CO2
P
(mole ▪ K) (0.97 atm)
8. Consider the reaction between 50.0 mL of liquid methyl alcohol, CH3OH (density = 0.850 g/mL), and
22.8 L of O2 at 27 oC and a pressure of 2.00 atm. The products of the reaction are CO2(g) and H2O(g).
Calculate the number of moles of H2O formed if the reaction goes to completion.
2 CH3OH
50.0 mL CH3OH
+
0.850 g CH3OH
1 mL CH3OH
3 O2

2 CO2
1 mole CH3OH
32.05 g CH3OH
+
4 mole H2O
2 mole CH3OH
n = PV = (2.00 atm) (22.8 L) (mole ▪ K) = 1.85 moles O2
RT
(0.0821 L ▪ atm) (300. K)
1.85 moles O2
4 mole H2O
3 moles O2
= 2.47 mole H2O
4 H2O
= 2.65 mole H2O
9. 4.40 gram sample of CO2 (s) is allowed to sublime in a 1.00 liter balloon that is at 300 K and contains
3.43 moles of Oxygen. What will be the final pressure in the balloon?
4.40 g CO2
1 moles CO2
44.01 g CO2
= 0.100 moles CO2
moles O2
+
moles CO2
3.43 mole
+
0.100 mole = 3.53 mole
P = nRT
V
=
= Total moles
(3.53 moles) (0.0821 L ▪ atm) (300 K)
(mole ▪ K) (1.00 L)
= 90 atm
10. A helium balloon has a volume of 2.30 L at 23.5 oC and a pressure of 1.00 atm at sea level. The
balloon is released and floats upward. At a certain height the atmospheric pressure is 0.810 atm
and the temperature is 12.0 oC. Calculate the volume of the balloon.
V2 = P1V1T2
T1P2
=
(1.00 atm) (2.30 L) (285 K)
(297 K) (0.810 atm)
=
2.72 L
11. A mole of a gas at STP has a volume of 22.4 L. If the volume increases to 29.1 L and the pressure
is changed to 813 torr, calculate the new temperature.
T2 = P2V2 T1 = (1.07 atm) (29.1 L) (273 K)
P1V1
(1.00 atm) (22.4L)
=
379 K
12. A sample of helium gas occupies 12.4 L at 23 oC and 0.956 atm. What volume will it occupy at
40 oC and 0.956 atm?
V2 = V1T2 = (12.4 L) (310 K) = 13 L
T1
(296 K)
13. C2H4 reacts with O2 according to the following unbalanced equation: C2H4(g) + O2(g)  CO2(g) + H2O(g)
What volume of oxygen at STP is needed to react with 1.50 mol of C2H4?
C2H4(g)
1.50 moles C2H4
3 mole O2
1 mole C2H4
+
3 O2(g)

2 CO2(g)
22.4 L O2
1 moles O2
+
2 H2O(g)
= 101 L O2
14. Calcium hydride combines with water according to the following unbalanced equation
CaH2(s) + H2O(l)  H2(g) + Ca(OH)2(s). Beginning with 84.0 g of CaH2 and 36.0 g of H2O, what
volume of H2 will be produced at a temperature of 273 K and a pressure of 1520 torr?
CaH2(s)
+
2 H2O(l)

2 H2(g)
+
Ca(OH)2(s)
84.0 g CaH2
1 mole CaH2
42.10 g CaH2
2 mole H2
1 moles CaH2
= 3.99 mole H2
36.0 g H2O
1 mole H2O
18.02 g H2O
2 mole H2
2 moles H2O
= 2.00 moles H2
V = nRT = (2.00 mole) (0.0821 L ▪ atm) (273 K) = 22.4 L H2
P
(mole ▪ K) (2.00 atm)