Question

Ethane burns in oxygen to form ${CO_2}$ and ${H_2O}$ according to the equation:
\[2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O\]
If 1250cc of oxygen is burnt with 300cc of ethane. Calculate:
a) The volume of ${CO_2}$ formed.
b) The volume of unused ${O_2}$.

Answer 1

Hint: Ethane is an organic chemical compound with chemical formula \[{C_2}{H_6}\]. And it has the double bond between carbon and hydrogen.

Complete answer:
Ethane is colorless and odorless gas. Its chief use is as feedstock for ethylene production.
${C{H_3}CO{O^ - } \to C{H_3}^\bullet + C{O_2} + {e^ - }}$
${C{H_3}^\bullet + ^\bullet C{H_3} \to {C_2}{H_6}}$
Combustion occurs by a complex series of free-radical reactions. An important series of reaction in ethane combustion is the combination of an ethyl radical with oxygen, and the subsequent breakup of the resulting peroxide into methoxy and hydroxyl radicals.
${{C_2}{H_5}^\bullet + {O_2} \to {C_2}{H_5}OO^\bullet}$
${{C_2}{H_5}OO^\bullet + HR \to {C_2}{H_5}OOH + ^\bullet R}$
${{C_2}{H_5}OOH \to {C_2}{H_5}O^\bullet + ^\bullet OH}$
Rotating a molecular substructure about a twistable bond usually requires energy. The minimum energy to produce a bond rotation is called the rotational barrier. Ethane gives a classic, simple example of such a rotational barrier. Sometimes, it’s called an ethane barrier.
The chief use of ethane is the production of ethane by steam cracking.
Oxygen is a chemical element with atomic number 8. It is a member of the chalcogen group in the periodic table, a highly reactive nonmetal and an oxidizing agent that readily forms oxides with most elements as well as with other elements present in the periodic table.

In above reaction \[2{C_2}{H_6} + 7{O_2} \to 4C{O_2} + 6{H_2}O\] we used 2 molecules of ethane and 7 molecules of oxygen and we calculate the formula of carbon dioxide which having 4 molecule as a result. So we use 1250cc of oxygen and 300cc of ethane so 300cc of ethane react with \[\dfrac{300 \times 7}{2}\] = 1050cc .

a) 2 molecule of ethane produced 4 molecule of carbon dioxide, so volume of \[C{O_2}\] is
\[\dfrac{300 \times 4}{2}\] = \[600cc\] of \[C{O_2}\]

b) Volume of unused oxygen is = 1250cc – 1050cc = 200cc

Note: We calculate volume by multiplying given volume to their molecule used in the reaction. And after that to find the volume of unused oxygen we have to subtract from the given volume.