Download More Stoichiometry (gas

CHAPTER FOUR
More Stoichiometry (Gas related problems)
4.88
This is an extension of an ideal gas law calculation involving molar mass. 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 ¹
PV
©
=
= 0.00168 mol
n =
L ⋅ atm ·
RT
§
¨ 0.0821 mol ⋅ K ¸ (77 + 273)K
©
¹
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 CaF2.
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
mol) 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.
4.89
We can calculate the moles of M reacted, and the moles of H2 gas produced. By comparing the number
of moles of M reacted to the number of moles H2 produced, we can determine the mole ratio in the
balanced equation.
Step 1: First let’s calculate the moles of the metal (M) reacted.
mol M = 0.225 g M ×
1 mol M
= 8.33 × 10−3 mol M
27.0 g M
Step 2: Solve the ideal gas equation algebraically for nH 2 .
substituting the known quantities into the equation.
Then, calculate the moles of H2 by
1 atm
= 0.975 atm
760 mmHg
P = 741 mmHg ×
T = 17° + 273° = 290 K
nH 2 =
nH 2 =
PVH 2
RT
FG
H
(0.975 atm)(0.303 L)
= 1.24 × 10−2 mol H2
L ⋅ atm
0.0821
(290 K)
mol ⋅ K
IJ
K
Step 3: Compare the number moles of H2 produced to the number of moles of M reacted.
1.24 × 10−2 mol H 2
8.33 × 10−3 mol M
≈ 1.5
This means that the mole ratio of H2 to M is 1.5 : 1.
Step 4: We can now write the balanced equation since we know the mole ratio between H2 and M.
The unbalanced equation is:
→ 1.5 H2 (g) + MxCly (aq)
M (s) + HCl (aq) 
We have 3 atoms of H on the products side of the reaction, so a 3 must be placed in front of
HCl. The ratio of M to Cl on the reactants side is now 1 : 3. Therefore the formula of the metal
chloride must be MCl3. The balanced equation is:
M (s) + 3 HCl (aq) 
→ 1.5 H2 (g) + MCl3 (aq)
From the formula of the metal chloride, we determine that the charge of the metal is +3.
Therefore, the formula of the metal oxide and the metal sulfate are M2O3 and M2(SO4)3,
respectively.
4.90
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 NH4Cl formed is:
? g NH4Cl = 2.00 mol HCl ×
1 mol NH 4 Cl
53.49 g NH 4 Cl
×
= 107 g NH4Cl
1 mol HCl
1 mol NH 4 Cl
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 =
4.91
nNH3 RT
P
L ⋅ atm ·
§
(2.29 mol) ¨ 0.0821
(14 + 273)K
mol ⋅ K ¸¹
©
=
= 54.5 L NH3
§
1 atm ·
752
mmHg
×
¨
¸
760 mmHg ¹
©
From the moles of CO2 produced, we can calculate the amount of calcium carbonate that must have
reacted. We can then determine the percent by mass of CaCO3 in the 3.00 g sample.
The balanced equation is:
→ CO2 (g) + CaCl2 (aq) + H2O (l)
CaCO3 (s) + 2 HCl (aq) 
The moles of CO2 produced can be calculated using the ideal gas equation.
nCO2 =
nCO 2 =
PVCO2
RT
FG 792 mmHg × 1 atm IJ (0.656 L)
760 mmHg K
H
⋅
L
atm
FG 0.0821
IJ (20 + 273 K)
H
mol ⋅ K K
= 2.84 × 10−2 mol CO2
The balanced equation shows a 1:1 mole ratio between CO2 and CaCO3. Therefore, 2.84 × 10−2 mol of
CaCO3 must have reacted.
? g CaCO3 reacted = 2.84 × 10−2 mol CaCO3 ×
1001
. g CaCO 3
= 2.84 g CaCO3
1 mol CaCO 3
The percent by mass of the CaCO3 sample is:
% CaCO3 =
2.84 g
× 100% = 94.7%
3.00 g
Assumption: The impurity (or impurities) must not react with HCl to produce CO2 gas.
4.92
The balanced equation is:
H2 (g) + Cl2 (g) 
→ 2 HCl (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 H2 reacted = 5.6 L H2 ×
1 mol H 2
= 0.25 mol H2
22.41 L H 2
The mass of HCl produced is:
? g HCl = 0.25 mol H2 ×
4.93
2 mol HCl
36.46 g HCl
×
= 18 g HCl
1 mol H 2
1 mol HCl
The balanced equation is:
→ 2 CO2 (g) + 3 H2O (l)
C2H5OH (l) + 3 O2 (g) 
The moles of O2 needed to react with 227 g ethanol are:
227 g C2H5OH ×
1 mol C 2 H 5OH
3 mol O 2
×
= 14.8 mol O2
46.07 g C 2 H 5OH
1 mol C 2 H 5OH
14.8 mol of O2 corresponds to a volume of:
VO 2 =
nO 2 RT
P
=
FG
H
FG
H
IJ
K
L ⋅ atm
(35 + 273 K)
mol ⋅ K
= 3.60 × 102 L O2
1 atm
790 mmHg ×
760 mmHg
(14.8 mol O 2 ) 0.0821
IJ
K
Since air is 21.0 percent O2 by volume, we can write:
Vair = VO 2
FG 100% air IJ
H 21% O K
2
= (3.60 × 102 L O 2 )
FG 100% air IJ
H 21% O K
2
= 1.71 × 103 L air