Irvington High School  AP Chemistry
Mr. Markic
Name _________________________________
Number ___ Date ___/___/___
5  Gases
Gas Stoichiometry
1.
Consider the formation of nitrogen dioxide from nitric oxide and oxygen:
2NO(g) + O2(g) → 2NO2(g)
If 9.0 L of NO are reacted with excess O2 at STP , what is the volume in liters of the NO2 produced?
9.0 L NO 
2.
2 volumes NO2
 9.0 L NO2
2 volumes NO
When coal is burned, the sulfur present in coal is converted to sulfur dioxide (SO2), which is
responsible for the acid rain phenomenon.
S(s) + O2(g) → SO2(g)
If 2.54 kg of S are reacted with oxygen, calculate the volume of SO2 gas (in mL) formed at 30.5C
and 1.12 atm.
(2.54  103 g S) 
nRT
V 

P
3.
1 mol SO2
1 mol S

 79.2 mol SO 2
32.07 g S
1 mol S
L  atm 

(79.2 mol)  0.0821
(303.5 K)
mol  K 

 1.76  103 L  1.76  106 mL SO2
1.12 atm
A compound of P and F was analyzed as follows: Heating 0.2324 g of the compound in a 378cm3 container turned all of it to gas, which had a pressure of 97.3 mmHg at 77C. Then the gas
was mixed with calcium chloride solution, which turned all of the F to 0.2631 g of CaF2.
Determine the molecular formula of the compound.
If you determine the molar mass of the gas, you will be able to determine the molecular formula from the empirical
formula. First, let’s calculate the molar mass of the compound.

1 atm 
 97.3 mmHg 
 (0.378 L)
760 mmHg 
PV
n 
 
 0.00168 mol
L  atm 
RT

 0.0821 mol  K  (77  273)K


Solving for the molar mass:
M 
mass (in g)
0.2324 g

 138 g/mol
mol
0.00168 mol
To calculate the empirical formula, first we need to find the mass of F in 0.2631 g of CaF 2.
0.2631 g CaF2 
1 mol CaF2
2 mol F
19.00 g F


 0.1280 g F
78.08 g CaF2 1 mol CaF2
1 mol F
Since the compound only contains P and F, the mass of P in the 0.2324 g sample is:
0.2324 g  0.1280 g  0.1044 g P
Now, we can convert masses of P and F to moles of each substance.
? mol P  0.1044 g P 
1 mol P
 0.003371 mol P
30.97 g P
? mol F  0.1280 g F 
1 mol F
 0.006737 mol F
19.00 g F
Thus, we arrive at the formula P0.003371F0.006737. Dividing by the smallest number of moles (0.003371 mole) gives the
empirical formula PF2.
To determine the molecular formula, divide the molar mass by the empirical mass.
molar mass
138 g

 2
empirical molar mass
68.97 g
Hence, the molecular formula is (PF2)2 or P2F4.
4. What is the mass of the solid NH4Cl formed when 73.0 g of NH3 are mixed with an equal mass of
HCl? What is the molar volume of the gas remaining, measured at 14.0C and 752 mmHg? What
gas is it?
The balanced equation for the reaction is: NH3(g)  HCl(g) 
 NH4Cl(s)
First, we must determine which of the two reactants is the limiting reagent. We find the number of moles of each
reactant.
? mol NH3  73.0 g NH3 
1 mol NH3
 4.29 mol NH3
17.03 g NH3
? mol HCl  73.0 g HCl 
1 mol HCl
 2.00 mol HCl
36.46 g HCl
Since NH3 and HCl react in a 1:1 mole ratio, HCl is the limiting reagent. The mass of NH 4Cl formed is:
? g NH4Cl  2.00 mol HCl 
1 mol NH 4 Cl 53.49 g NH 4Cl

 107 g NH 4Cl
1 mol HCl
1 mol NH 4Cl
The gas remaining is ammonia, NH3. The number of moles of NH3 remaining is (4.29  2.00) mol  2.29 mol NH3. The
volume of NH3 gas is:
VNH 3 
nNH3 RT
P
L  atm 

(2.29 mol)  0.0821
(14  273)K
mol  K 


 54.5 L NH 3

1 atm 
752
mmHg



760 mmHg 

5. Calculate the mass in grams of hydrogen chloride produced when 5.6 L of molecular hydrogen
measured at STP react with an excess of molecular chlorine gas.
The balanced equation is:
H2(g)  Cl2(g) 
 2HCl(g)
At STP, 1 mole of an ideal gas occupies a volume of 22.41 L. We can use this as a conversion factor to find the moles of
H2 reacted. Then, we can calculate the mass of HCl produced.
? mol H 2 reacted  5.6 L H 2 
1 mol H2
 0.25 mol H 2
22.41 L H 2
The mass of HCl produced is:
? g HCl  0.25 mol H2 
2 mol HCl 36.46 g HCl

 18 g HCl
1 mol H2
1 mol HCl