Gas Laws (Part II): Gay-Lussac's Law & Avogadro's Law

The Gay-Lussac's Law of Combining Volumes states that when gases react, they do so in volumes, which are in simple ratio to one another and to the volume of the product, if any; provided temperature and pressure remain constant. It applies to only gases, which means that solid and liquid reactants and products are always ignored when applying this law.

For instance, hydrogen burns in oxygen at 100°C to form steam according to the equation:

       2H2(g) + O2(g) ---> 2H2O(g)
         2mol      1mol           2mol
         2vol        1vol            2vol
         2cm^3    1cm^3        2cm^3

From the above, it implies that at 100°C, when water is in its gaseous state, 2 volumes of hydrogen gas (dm^3 or cm^3) will combine with 1 volume of oxygen gas to form 2 volumes of steam, to give a simple mole ratio of 2 : 1 : 2. Therefore, 50cm^3 of hydrogen will need 25cm^3 of oxygen to produce 50cm^3 of steam. Similarly, 15cm^3 of oxygen will require 30cm^3 of hydrogen to form 30cm^3 of steam. In other words, if you start with (2 + 1 = 3) volumes of gases (hydrogen + oxygen); after the reaction, you will be left with 2 volumes of residual gas (steam).

Now, what will happen if this same reaction is carried out at a temperature lower than 100°C, say 50°C. The product will not be steam but water. This is because at the reaction temperature of 50°C, water is liquid, and not gaseous. So, the equation of reaction becomes:

        2H2(g) + O2(g) ---> 2H2O(l)
         2mol      1mol           2mol
         2vol        1vol                -
         2cm^3    1cm^3            -

From the analysis, it is only hydrogen and oxygen that contribute to the gaseous volume of the reaction, the water [H2O(l)] has no effect. Therefore, if you start with (2 + 1 = 3) volumes of gases (hydrogen + oxygen); after the reaction, you will be left with zero or no volume of residual gas.
According to Gay-Lussac, the gaseous reactants have no relationship with the liquid product, in terms of gas volumes.

Examples

Question 1
65cm^3 of hydrogen are sparked with 30cm^3 of oxygen at 100°C and 1atm. Calculate the total volume of residual gases at:
a) 100°C
b) room temperature

Answer
a)                            2H2(g) + O2(g) ---> 2H2O(g)
                                2vol           1vol             2vol
Before sparking     65cm^3    30cm^3       -
On sparking            60cm^3   30cm^3       60cm^3
After sparking          5cm^3      0cm^3       60cm^3

Therefore, volume of residual gas left at 100°C will be
= 5cm^3 of hydrogen + 60cm^3 of steam
= 65cm^3 of gases

Analysis
From the stoichiometry, 2 volumes of hydrogen will burn in 1 volume of oxygen to produce 2 volumes of steam, right? Going by the volume of gases available, what volume of oxygen will we need to completely spark 65cm^3 of hydrogen? That will be 1/2 of 65 = 32.5cm^3, isn't it? But the volume of oxygen available is only 30cm^3, which means it is not enough to completely spark the available hydrogen. So, we make use of the available oxygen, which implies that only 30 x 2 = 60cm^3 of hydrogen will react, while 65 - 60 = 5cm^3 of hydrogen will be left unreactive.

Since the oxygen is not enough to completely burn the hydrogen gas, it means it was used up and nothing was left unreactive, right? Also, from the stoichiometry, 30cm^3 of oxygen will produce 60cm^3 of steam, won't it? This implies that since all the oxygen reacted with hydrogen, it means 60cm^3 of steam was produced. Recall that no steam was produced before the sparking. Therefore, 60cm^3 of steam will be left as a residual gas, in addition to the 5cm^3 of unreactive hydrogen to give us a total of 60 + 5 = 65cm^3 of residual (left-over) gas.

b) Equation of reaction:
                               2H2(g) + O2(g) ---> 2H2O(l)
                                2vol      1vol             -
Before sparking     65cm^3    30cm^3       -
On sparking          60cm^3    30cm^3       -
After sparking          5cm^3      0cm^3        -

Therefore, volume of residual gas at room temperature will be
= 5cm^3 of hydrogen

Analysis
The reaction is similar to the one above, except that it occurs at room temperature, and at this temperature, water is a liquid and not gas. [Note that room temperature is the normal temperature of the room (without air-conditioner or heater) at a particular time]. In that case, it is only the reactants that will account for the residual gases if any. Since 5cm^3 of hydrogen is the only volume of gas left after the reaction, it therefore, accounts for the volume of residual gas as 5cm^3.

Question 2
60cm^3 of a mixture of carbon (II) oxide and hydrogen yielded 20cm^3 of carbon (IV) oxide after explosion with excess of oxygen. What is the percentage of carbon (II) oxide in the mixture?

Answer
       2CO(g) + O2(g) ---> 2CO2(g)
        2vol        1vol              2vol

From the question, no concrete information is provided for oxygen, so will have to ignore it and focus on CO and CO2.

No information is provided on the actual volume of CO that reacted with the excess oxygen, but we can use the volume of CO2 produced to get that as shown below:

If 2vol of CO2 = 2vol of CO
Then, 20cm^3 of CO2 = 20cm^3 of CO.

The percentage of CO in the mixture is given by:
(Volume of CO that reacted/Volume of mixture) x 100
= (20/60) x 100
= 0.3333 x 100
= 33.33%

It is important to note that since the product of the explosion between the mixture and oxygen was CO2, it means that it was only CO that took part in the reaction.

Question 3
25cm^3 of hydrogen was sparked with 20cm^3 of oxygen at atmospheric pressure. What is the total volume of the residual gas at room temperature?

Answer
Temperature of reaction = room temperature (>0°C and <100°C), this means that the product was water and not steam.
                               2H2(g) + O2(g) ---> 2H2O(l)
                                2vol           1vol             2vol
Before sparking     25cm^3    20cm^3        -
On sparking            25cm^3   12.5cm^3     -
After sparking          0cm^3      7.5cm^3      -

Therefore, total volume of residual gas left at room temperature was
= 7.5cm^3 of oxygen

Analysis
From the stoichiometry, 20 x 2 = 40cm^3 of hydrogen will be required to completely react with the 20cm^3 of oxygen; but the volume of hydrogen available is just 25cm^3, and it will require 25/2 = 12.5cm^3 of oxygen to completely react as shown above.

Question 4
20cm^3 of carbon (II) oxide is mixed and sparked with 80cm^3 of air containing 21% oxygen. If all the volumes are measured at s.t.p., calculate the volume of the resulting gases.

Answer
Volume of air = 80cm^3
Percentage of oxygen in air = 21%
Actual volume of oxygen = (21/100) x 80
= 16.80cm^3
Remaining components of air that do not react = Volume of air - Volume of oxygen
= 80 - 16.8
= 63.2cm^3 (made up of nitrogen, carbon (IV) oxide, water vapour and noble gases)

At s.t.p., carbon (II) oxide, oxygen and carbon (IV) oxide are all gases. So, the equation of reaction is:
      
                                2CO(g) + O2(g) ---> 2CO2(g)
                                 2vol          1vol           2vol
Before sparking     20cm^3    16.8cm^3        -
On sparking            20cm^3   10cm^3       20cm^3
After sparking          0cm^3      6.8cm^3      20cm^3

Total volume of resulting gas = Volume of excess oxygen + Volume of CO2 produced + Volume of unreactive components of air
= 6.8cm^3 + 20cm^3 + 63.2cm^3
= 90cm^3

Question 5
2H2(g) + O2(g) ---> 2H2O(g)
In the reaction above, what volume of hydrogen would be left over when 300cm^3 of oxygen and 1000cm^3 of hydrogen are exploded in a sealed tube?

Answer
                                 2H2(g) + O2(g) ---> 2H2O(g)
                                   2vol          1vol            2vol
Before explosion  1000cm^3   300cm^3        -
On explosion        600cm^3    300cm^3       600cm^3
After explosion     400cm^3       0cm^3        600cm^3

Therefore, the volume of hydrogen that would be left = 400cm^3

Question 6
35cm^3 of hydrogen was sparked with 12cm^3 of oxygen at 110°C and 760mmHg to produce steam. What percentage of the total volume of gas left after the reaction is hydrogen?

Answer
                                   2H2(g) + O2(g) ---> 2H2O(g)
                                   2vol          1vol              2vol
Before sparking        35cm^3   12cm^3          -
On sparking              24cm^3    12cm^3       24cm^3
After sparking          11cm^3       0cm^3        24cm^3

Total volume of residual gas = Volume of unreactive hydrogen + Volume of steam produced
= 11cm^3 + 24cm^3
= 35cm^3

Percentage of hydrogen in residual gas
= (Volume of excess (unreactive) hydrogen/Total volume of residual gas) x 100
= (11/35) x 100
= 0.3143 x 100
= 31.43%

Avogadro's Law
This law states that equal volumes of all gases contain the same number of molecules at the same temperature and pressure.  It is related to Gay-Lussac's Law, as its application shows that the volume of gases can be converted into number of molecules at constant temperature and pressure. It implies that if we know the equal volumes of two gases A and B, and the number of molecules of one of the gases, we can easily determine the number of molecules of the other gas.

For instance, when hydrogen gas combines with oxygen gas to form steam according to the equation:

                               2H2(g) + O2(g) ---> 2H2O(g)
Mole ratio               2         :       1         :        2
Gay-Lussac's          2vol   :      1vol     :      2vol
Avogadro's              2        :       1         :        2
                          molecules   molecule     molecules

Similarly, the reaction between nitrogen and hydrogen to form ammonia can be expressed as:

                                N2(g) + 3H2(g) ---> 2NH3(g)
Mole ratio               1         :       3         :        2
Gay-Lussac's          1vol   :      3vol     :      2vol
Avogadro's              1        :       3         :        2
                          molecule   molecules     molecules

Also, the reaction between carbon (II) oxide and oxygen to produce carbon (IV) oxide:

                               2CO(g) + O2(g) ---> 2CO2(g)
Mole ratio               2         :       1         :        2
Gay-Lussac's          2vol   :      1vol     :      2vol
Avogadro's              2        :       1         :        2
                          molecules   molecule     molecules

Hydrogen reacts with fluorine to form hydrogen fluoride according to the equation:

                                 H2(g) + F2(g) ---> 2HF(g)
Mole ratio               1         :       1         :        2
Gay-Lussac's          1vol   :      1vol     :      2vol
Avogadro's              1        :       1         :        2
                          molecule   molecule     molecules

In the above equations, we can see that all the volumes of the gases are converted to molecules, based on the assumption that
1 mol of a gas = 1 vol of the gas = 1 molecule of the gas, at constant temperature and pressure.

Examples

Question 7
8g of methane, CH4 occupies 11.2dm^3 at s.t.p. What volume would 22g of propane, C3H8 occupy under the same condition?

Answer
Convert the given masses to moles
Molar mass of CH4 = (1*C) + (4*H)
                                   = (1*12) + (4*1)
                                   = 12 + 4
                                   = 16g/mol

Molar mass of C3H8 = (3*C) + (8*H)
                                     = (3*12) + (8*1)
                                     = 36 + 8
                                     = 44g/mol

Reacting mass of CH4 = 8g
Number of moles of CH4 = 8/16
                                            = 0.5mol

Reacting mass of C3H8 = 22g
Number of moles of C3H8 = 22/44
                                           = 0.5mol

If 0.5mol of CH4 occupies 11.2dm^3
Then, 0.5mol of C3H8 will also occupy 11.2dm^3

According to Avogadro's Law,since CH4 and C3H8 contain the same number of moles under the same condition, it implies that they will occupy the same volume.

Question 8
Calculate the volume of oxygen required to completely burn 20cm^3 of ethane (C2H6)

Answer
Equation of reaction:
             2C2H6(g) + 7O2(g) ---> 4CO2(g) + 6H2O(g)
Mole ratio    2       :    7             :     4             :   6
Vol ratio    2vol         7vol              4vol            6vol

From the above,
2cm^3 of C2H6 = 7cm^3 of oxygen
20cm^3 of C2H6 = x cm^3 of oxygen
                     2 * x = 20 * 7
                         2x = 140
                           x = 140/2
                              = 70cm^3 of oxygen

Question 9
What is the volume of propyne (C3H4) required to completely burn in 20cm^3 of oxygen?

Answer
Equation of reaction:
              C3H4(g) + 4O2(g) ---> 3CO2(g) + 2H2O(g)
Mole ratio    1       :    4             :     3             :   2
Vol ratio      1vol       4vol              3vol            2vol

From the stoichiometry,
1cm^3 of C3H8 = 4cm^3 of oxygen
x cm^3 of C3H8 = 20cm^3 of oxygen
                     4 * x = 20 * 1
                         4x = 20
                           x = 20/4
                              = 5cm^3 of propyne 

Do These

Question 1
20cm^3 of hydrogen gas are sparked with 20cm^3 of oxygen gas at 373K (100°C) and 1 atmosphere. The resulting mixture is cooled to 298K (25°C) and passed over calcium chloride. Calculate the volume of the residual gas.
A.  40cm^3   B.  20cm^3   C.  30cm^3   D.  10cm^3

Question 2
10cm^3 of hydrogen fluoride gas (HF) reacts with 5cm^3 of dinitrogen difluoride gas (N2F2) to form 10cm^3 of a single gas (NHF2).
a) Write the most likely equation of the reaction.
b) What volume of gas will be left if 50cm^3 of HF reacts with 40cm^3 of N2F2?

Question 3
Calculate the minimum volume of oxygen required for the complete combustion of a mixture of 10cm^3 of CO and 15cm^3 of H2.
A.  25.0cm^3  B.  12.5cm^3  C.  10.0cm^3   D.  5.0cm^3

Question 4
20 volumes of a mixture of hydrogen and carbon (II) oxide were mixed with 13 volumes of oxygen and the mixture exploded. On cooling, the resulting volume was found to be 9. Determine the composition of the mixture.

Question 5
One mole of propane is mixed with five moles of oxygen. The mixture is ignited and the propane burns completely. What is the volume of the products at s.t.p? [H = 1, C = 12, O = 16, G.M.V = 22.4dm^3 per mol]

Question 6
50cm^3 of carbon (II) oxide was exploded with 150cm^3 of air containing 20% oxygen by volume. Which of the reactants was in excess?
A. Carbon (II) oxide  B. Oxygen  C. Carbon (IV) oxide  D. Nitrogen

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