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Stoichiometry Problems Involving Gases

Skills to develop

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 carbon-12, 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, n1, and P1 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:

ntotal = n1 + n2 + n3 + . . . + nn Since n = (V/RT) P, the number of moles of the ith component is related to its partial pressure in the same formula, ni = (V/RT) Pi, and ntotal = (V/RT) Ptotal. Therefore, Ptotal = P1 + P2 + P3 + . . . + Pn

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 nHCl = 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 + H2O   <== 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 5.0-L air sample containing H2S at STP is treated with a catalyst to promote the reaction,
H2S + O2 = H2O + S(solid). If 3.2 g of solid S was collected, calculate the volume percentage of H2S in the origional sample.

Solution:

         1 mol H2S
 3.2 g S --------- = 0.10 mol H2S
         32 g S

 V_H2S = 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 H2O and S, H2S + SO2 = H2O + S(solid), unbalanced. If 6.0 L of H2S gas at 750 torr produced 3.2 g of sulfur, calculate the temperature in C.

Solution
Balanced reaction:

 2 H2S + SO2 = 2 H2O + 3 S(solid),
 2 mol                3*32 = 96 g

         2 mol H2S
 3.2 g S ---------- = 0.067 mol H2S;
          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 AgNO3 solution is treated with excess amount of HI gas to give 2.35 g of AgI, calculate the concentration of AgNO3solution.

Solution

            1 mol Ag+    1 mol AgNO3
2.35 g AgI ----------- --------------
           234.8 g AgI   1 mol Ag+

        = 0.010 mol AgNO3

[AgNO3] = 0.01 mol AgNO3 / 0.050 L
        = 0.20 M AgNO3
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 3.66-g sample containing Zn (at.wt. 65.4) and Mg (24.3) reacted with a dilute acid to produce 2.470 L H2 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 H2O = 2 Na(OH) + H2(g)
Ca + H2O = Ca(OH) + H2(g)

Let x be the mass of Na, then (2.00-x) 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 H2
----------
1 mol Ca
  = 1.164 L H2 * 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 MgCl2.

Also compare with analyses making use of the reaction
Ba2+(aq) + SO42-(aq) = BaSO4 (s)
where the anion SO42-(aq) comes from the disolution of two sulfate salts.

This example is very similar to Example 7.

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