
In the combustion of acetylene, $\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{{\text{O}}_{\text{2}}}\to \text{4C}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}$.
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
Answer
547.8k+ views
Hint: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,
$\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{{\text{O}}_{\text{2}}}\to \text{4C}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}$
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 = $\left[ \left( 12\times 2 \right)+\left( 1\times 2 \right) \right]$= 26 grams
One mole of oxygen = the molecular weight oxygen = 32 grams, 5 moles of oxygen = $5\times 32$=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 $\dfrac{160}{2}=80$grams 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.
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,
$\text{2}{{\text{C}}_{\text{2}}}{{\text{H}}_{\text{2}}}\text{ + 5}{{\text{O}}_{\text{2}}}\to \text{4C}{{\text{O}}_{\text{2}}}\text{+ 2}{{\text{H}}_{\text{2}}}\text{O}$
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 = $\left[ \left( 12\times 2 \right)+\left( 1\times 2 \right) \right]$= 26 grams
One mole of oxygen = the molecular weight oxygen = 32 grams, 5 moles of oxygen = $5\times 32$=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 $\dfrac{160}{2}=80$grams 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.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Class 11 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

