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Stoichiometry Problems Involving Gases
Skills to develop
 Apply gas laws to solve stoichiometry problems.
 Apply principles of stoichiometry to calculate properties of gases.
The quantitative relationship of reactants and products is called
stoichiometry. Stoichiometric problems require you to
calculate the amounts of reactants required for certain amounts
of products, or amounts of products produced from certain amounts
of reactants. If, in a chemical reaction, one or more reactants
or products are gases, gas laws must be considered for the calculation.
Usually, the applications of the
ideal gas law
give results within 5% precision.
On this page, we review several important concepts that are helpful
for solving Stoichiometry Problems Involving Gases.
The Mole Concept
The mole concept is the key to both stoichiometry and gas laws.
A mole is a definite
amount of substance.
Mole is a unit based on the number of identities (i.e. atoms, molecules, ions,
or particles). A mole of anything has the same number of identities as
the number of atoms in exactly 12 grams of carbon12, the most abundant
isotope of carbon.
Law of Partial Pressure
For gases, the partial pressure of a component is the same as if the
component is by itself in the container. The total pressure is the sum
of all partial pressures of components. This is Dalton's law of partial
pressures, which is one of the
ABCD gas laws.
In the following discussion, n_{1}, and P_{1}
represent number of moles and partial pressure of the component 1
respectively. A similar notation can be given to components 2, 3, 4, etc.
When several components are present in a container, the total number of
moles is the sum of number of moles of the components:
n_{total} = n_{1} + n_{2}
+ n_{3} + . . . + n_{n}
Since n = (V/RT) P, the number of moles of the ith component
is related to its partial pressure in the same formula,
n_{i} = (V/RT) P_{i},
and
n_{total} = (V/RT) P_{total}.
Therefore,
P_{total} = P_{1} + P_{2}
+ P_{3} + . . . + P_{n}
Stoichiometry and Gas Laws
Stoichiometry is the theme of the previous block of modules, and the
ideal gas law is the theme of this block of modules. These subjects are
related. Be prepared to solve problems requiring concepts
or principles of stoichiometry and gases.
For example, we can calculate the number of moles from certain volume,
temperature and pressure of a HCl gas. When n moles dissolved
in V L solution, its concentration is n/V M.
Three examples are given to illustrate some calculations of stoichiometry
involving gas laws. More are given in question form for you to practice.
Example 1
If 500 mL of HCl gas at 300 K and 100 kPa dissolve in 100 mL of pure
water, what is the concentration? Data required: R value 8.314 kPa
L / (K mol).
Solution:
0.50 L * 100 kPa
n_HCl = 
(8.314 kPa L/(K mol) * 300 K)
= 0.02 mol
Concentration of HCl, [HCl]
[HCl] = 0.02 mol / 0.1 L = 0.2 mol/L.
Discussion
Note that R = 0.08205 L atm /(K mol) will not be suitable in this case.
If you have difficulty, review
Solutions.
Example 2
If 500 mL of HCl gas at 300 K and 100 kPa dissolved in pure water requires
12.50 mL of the NaOH solution to neutralize in a titration experiment,
what is the concentration of the NaOH solution?
Solution:
Solution in Example 1 showed n_{HCl} = 0.02 mol.
From the titration experiment, we can conclude that there were 0.02 moles
of NaOH in 12.50 mL. Thus,
[NaOH] = 0.02 mol / 0.0125 L = 1.60 mol/L
Discussion:
Think in terms of reaction,
HCl + NaOH = NaCl + H_{2}O <== Reaction
0.02 mol 0.02 mol <== Quantities reacted
Note: that 0.02 mol of NaOH is in 0.0125 mL solution.
Example 3:
A 5.0L air sample containing H_{2}S at STP is treated with a catalyst
to promote the reaction,
H_{2}S + O_{2} = H_{2}O + S(solid).
If 3.2 g of solid S was collected, calculate the volume percentage of H_{2}S
in the origional sample.
Solution:
1 mol H_{2}S
3.2 g S  = 0.10 mol H_{2}S
32 g S
V_H_{2}S = 0.10 mol * 22.4 L/mol
= 2.24 L
Volume % = 2.25 L / 5.0 L
= 0.45
= 45 %
Discussion:
Data required: Atomic mass: H, 1; O, 16; S, 32. R = 0.08205
L atm /(K mol) is now suitable R values or molar volume at STP (22.4
L/mol)
The volume percentage is also the mole percentage, but not the weight
percentage.
Example 4:
Hydrogen sulfide reacts with sulfur dioxide to give H_{2}O and S,
H_{2}S + SO_{2} = H_{2}O + S(solid), unbalanced.
If 6.0 L of H_{2}S gas at 750 torr produced 3.2 g of sulfur, calculate
the temperature in C.
Solution
Balanced reaction:
2 H_{2}S + SO_{2} = 2 H_{2}O + 3 S(solid),
2 mol 3*32 = 96 g
2 mol H_{2}S
3.2 g S  = 0.067 mol H_{2}S;
96 g S
P = 750/760 = 0.987 atm
PV 0.987 atm * 6 L
T =  = 
n R 0.067 mol * 0.08205 atm L /(mol K)
= 1085 K
= 812°C
Discussion:
Atomic mass: H, 1.0; O, 16.0; S, 32.0. R = 0.08205 L atm /(K mol)
is OK but watch units used for pressure.
Example 5:
When 50.0 mL of AgNO_{3} solution is treated with excess amount
of HI gas to give 2.35 g of AgI, calculate the concentration of
AgNO_{3}solution.
Solution
1 mol Ag^{+} 1 mol AgNO_{3}
2.35 g AgI  
234.8 g AgI 1 mol Ag^{+}
= 0.010 mol AgNO_{3}
[AgNO_{3}] = 0.01 mol AgNO_{3} / 0.050 L
= 0.20 M AgNO_{3}
Discussion
A gas is involved, but there is no need to consider the gas law.
At. mass: Ag, 107.9; N, 14.0; O, 16.0; I, 126.9
Example 6:
What volume (L) will 0.20 mol HI occupy at 300 K and 100.0 kPa?
R = 8.314 kPa L / (K mol) = 0.08205 atm L / (mol K).
Solution
n RT
V = 
P
0.20 mol * 8.314 kPa L / (mol K) *300 K
= 
100 kPa
= 5 L,
Example 7:
A 3.66g sample containing Zn (at.wt. 65.4) and Mg (24.3) reacted with a
dilute acid to produce 2.470 L H_{2} gas at 101.0 kPa and 300 K.
Calculate the percentage of Zn in the sample.
Solution
The number of moles of gas produced is the number of moles of metals
in the sample. Once you know the number of moles, set up an equation
to give the number of moles of metal in the sample.
n = 101 kPa * 2.470 L / (8.3145 kPa L / (mol K) * 300 K)
= 0.100 mol
Let x be the mass of Zn, then the mass of Mg is 3.66  x g.
Thus, we have
x 3.66  x
 +  = 0.100 mole
65.4 24.3
Solving for x gives x = 1.96 g Zn,
and the weight percent = 100 * 1.96 / 3.66 = 53.6 %
Discussion
Find the mole percent of Zn in the sample.
# mol of Zn = 1.96/65.4 = 0.03 mol
# mol of Mg = 1.70/24.3 = 0.07 mol
mole percent = 100 * 0.03 / (0.03 + 0.07) = 30 %
Example 8
When 2.00 g mixture of Na and Ca reat with water, 1.164 L hydrogen was
produced at 300.0 K and 100.0 kPa. What is the percentage of Na in the
sample?
solution
2 Na + 2 H_{2}O = 2 Na(OH) + H_{2}(g)
Ca + H_{2}O = Ca(OH) + H_{2}(g)
Let x be the mass of Na, then (2.00x) is the mass of Ca.
We have the following relationship
x g  23.0 g/mol  1 mol H2  2 mol Na
 +
 (2.0  x) g Ca  40.1 g Ca/mol  1 mol H_{2}  1 mol Ca
 =
 1.164 L H_{2} * 100.0 kPa 
8.3145 kPa L mol^{1} K^{1} 300.0 K


Simplify to give
x  46.0  +  2  40.1  
 x  40.1  =  0.0467  all in mol

Multiply all terms by (40.1 * 46.0)
40.1 x + 2 * 46.0  46.0 x = 86.1
Simplify
5.9 x = 86.1  92.0 = 5.91
Thus,
Mass of Na = x = 1.0 g
Mass of Ca = 2.0  x = 1.0 g
Mass Percentage of Na = 100* (1 / 2.0) = 50%
Discussion
Mole of Na = 1/23 = 0.0435 mol
Mole percentage = (^{1}/_{23}) /
(^{1}/_{23} + ^{1}/_{40.1}) = 0.635 = 63.5%
Compare this example with gravimetric analyses using the reaction
Ag^{+}(aq) + Cl^{}(aq) = AgCl (s)
where Cl^{}(aq) comes from the disolution of two salts such as
NaCl and MgCl_{2}.
Also compare with analyses making use of the reaction
Ba^{2+}(aq) + SO_{4}^{2}(aq) = BaSO_{4} (s)
where the anion SO_{4}^{2}(aq) comes from the
disolution of two sulfate salts.
This example is very similar to Example 7.
Confidence Building Questions


If 100 mL of HCl gas at 300 K and 100 kPa dissolved in pure water requires
12.50 mL of the NaOH solution to neutralize in a titration experiment,
what is the concentration of the NaOH solution?






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