Answer
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Hint:The combustion reaction of acetylene with oxygen gives carbon dioxide and water. This reaction follows the law of constant proportions and hence using stoichiometry we can determine the amount of air required for the burning of acetylene completely.
Complete step by step answer:
Let us write down the combustion reaction for acetylene:
${C_2}{H_2} + \dfrac{5}{2}{O_2} \to 2C{O_2} + {H_2}O$
From the above balanced reaction we know that acetylene burns in presence of oxygen gas to give out carbon dioxide gas and water vapours.
So from this we know that $\dfrac{5}{2}$ moles of oxygen gas reacts with 1 mole of acetylene.
or, 1 mL of acetylene gas reacts stoichiometrically with $\dfrac{5}{2}$ mL of oxygen gas.
Hence, $200{m^3}$ of acetylene gas will need = $\dfrac{{5mL}}{{2 \times 1mL}} \times 200{m^3}$ of oxygen gas.
Solving the above question, we get the value for the amount of oxygen gas required for the complete combustion of acetylene gas, which will be : $500{m^3}$ .
It is given in the question, that we need air, which has one of its components as oxygen gas for the combustion of acetylene and not pure oxygen
So, we need to find the amount of air we will need which will give us $500{m^3}$ of oxygen gas.
In the question, it is mentioned that the air has $20\% $ oxygen by volume.
this means that in $100mL$ of air , $20mL$ of it is oxygen
In other words, $20mL$ of oxygen can be found in $100mL$ of air
hence, $500{m^3}$ of oxygen gas can be found in = $\dfrac{{100 \times 500}}{{20}}{m^3}$
Solving the above equation we get the value of amount of air needed for the complete combustion if acetylene as $2500{m^3}$
Hence Option 1 is correct.
Note:
The percentage of air given in the question is in the form of volume by volume, this is a popular method for determining composition of gases .
Acetylene gas is used for welding purposes as it is capable of producing the hottest flame .
Complete step by step answer:
Let us write down the combustion reaction for acetylene:
${C_2}{H_2} + \dfrac{5}{2}{O_2} \to 2C{O_2} + {H_2}O$
From the above balanced reaction we know that acetylene burns in presence of oxygen gas to give out carbon dioxide gas and water vapours.
So from this we know that $\dfrac{5}{2}$ moles of oxygen gas reacts with 1 mole of acetylene.
or, 1 mL of acetylene gas reacts stoichiometrically with $\dfrac{5}{2}$ mL of oxygen gas.
Hence, $200{m^3}$ of acetylene gas will need = $\dfrac{{5mL}}{{2 \times 1mL}} \times 200{m^3}$ of oxygen gas.
Solving the above question, we get the value for the amount of oxygen gas required for the complete combustion of acetylene gas, which will be : $500{m^3}$ .
It is given in the question, that we need air, which has one of its components as oxygen gas for the combustion of acetylene and not pure oxygen
So, we need to find the amount of air we will need which will give us $500{m^3}$ of oxygen gas.
In the question, it is mentioned that the air has $20\% $ oxygen by volume.
this means that in $100mL$ of air , $20mL$ of it is oxygen
In other words, $20mL$ of oxygen can be found in $100mL$ of air
hence, $500{m^3}$ of oxygen gas can be found in = $\dfrac{{100 \times 500}}{{20}}{m^3}$
Solving the above equation we get the value of amount of air needed for the complete combustion if acetylene as $2500{m^3}$
Hence Option 1 is correct.
Note:
The percentage of air given in the question is in the form of volume by volume, this is a popular method for determining composition of gases .
Acetylene gas is used for welding purposes as it is capable of producing the hottest flame .
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