This concept of reacting volumes may be easily applied to calculate excess and limiting reagents in gaseous reactions. |
Limiting gas volume
By inspection of the volumes reacting, it is a simple matter to determine the limiting reagent and which is in excess.
N2
|
+
|
3H2
|
![]() |
2NH3
|
1 volume
|
|
3 volumes
|
|
2 volumes
|
Experimental data
|
|
|||
300 cm3
|
|
300 cm3
|
|
???
|
excess
|
|
limiting reagent
|
![]() |
200cm3 ammonia
|
In the table above, the amounts of nitrogen and hydrogen provided initially are 300 cm3 of each.
But from the stoichoimetry of the equation we can see that 300cm3 of nitrogen gas would require 3 x that volume of hydrogen gas for complete reaction. The nitrogen is therefore in excess and only 100cm3 of the nitrogen can react with hydrogen.
The reaction proportions then are 100 cm3 nitrogen reacting with 300 cm3 hydrogen to produce 200 cm3 of ammonia.
Note: If only 100cm3 nitrogen reacts there is also 200 cm3 nitrogen left over at the end of the reaction.
Example: Calculate (i) the volume of carbon dioxide produced when 300 cm3 of methane burns in 1000 cm3 oxygen and (ii) the final volume of the reaction mixture. The equation for the reaction: CH4(g) + 2O2(g) The ratio of gas volumes is 1 volume of methane reacts with 2 volumes of oxygen producing 1 volume of carbon dioxide (and liquid water, no volume) There is 300 cm3 methane - this needs 2 x 300 cm3 = 600 cm3 oxygen to burn completely and produce 1 x 300 cm3 carbon dioxide = 300cm3 carbon dioxide During the course of the reaction 300 cm3 methane and 600 cm3 of oxygen is used up from an initial total of 300 cm3 of methane and 1000 cm3 oxygen. There is 400 cm3 of oxygen in excess therefore the final mixture contains 300 cm3 carbon dioxide + 400 cm3 of oxygen = 700 cm3 of gas |