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Question: In the combustion of acetylene, \(\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{...

In the combustion of acetylene, 2C2H2 + 5O2→4CO2+ 2H2O\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{{\text{O}}_{\text{2}}}\to \text{4C}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}.
The amount of oxygen required for the complete combustion of one mole of acetylene is:
A.40 g
B.120 g
C.80 g
D.64 g

Explanation

Solution

The process of combustion means oxidation of a substance by oxygen. This involves a rapid reaction between the oxygen molecules and results in the liberation of heat and light energy. We can calculate the amount using the stoichiometric coefficients of the equation.

Complete step by step answer:
The reaction for the combustion of acetylene is based on the mole concept and stoichiometry and as such, and as given in the above reaction,
2C2H2 + 5O2→4CO2+ 2H2O\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{{\text{O}}_{\text{2}}}\to \text{4C}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}
2 moles of acetylene combines with five moles of oxygen to form four moles of carbon dioxide and 2 moles of water vapour.
One mole of any compound = the molecular weight of that compound
Hence one mole of acetylene = the molecular weight of acetylene = [(12×2)+(1×2)]\left[ \left( 12\times 2 \right)+\left( 1\times 2 \right) \right]= 26 grams
One mole of oxygen = the molecular weight oxygen = 32 grams, 5 moles of oxygen = 5×325\times 32=160 grams.
Therefore, as per the stoichiometry of the reaction,
52 grams, i.e., 2 moles of acetylene is oxidized by 160 grams or five moles of oxygen
Therefore, 1 mole of acetylene is oxidized by 1602=80\dfrac{160}{2}=80grams of oxygen.

Therefore, the correct answer is option C.

Note:
The combination between the atoms to form compounds or the combination of the compounds to form new compounds are guided by certain laws called the laws of chemical combination. This problem is based on the law of Gay Lussac’s Law of combining volumes which states that when gases react together to form different products then they do so in simple whole number ratios under similar conditions of temperature and pressure.