
Amount of oxygen required for combustion of $1\,Kg$ mixture of butane and isobutane is:
A) $1.8\,kg$
B) $2.7\,kg$
C) $4.5\,kg$
D) $3.58\,kg$
Answer
558.6k+ views
Hint: Firstly try to calculate the molar mass of butane, by which you get to know that amount of butane reacts with how much amount of oxygen by putting molar mass of oxygen. Then solve for one gram of butane, you have to divide grams of oxygen with molar mass of butane for one gram. Now then multiply the amount of oxygen you get with $1000\,g$ because $1Kg = 1000\,g$
Complete step-by-step answer:
We have formula for butane as ${C_4}{H_{10}}$ and for isobutane the formula is same, the different is that in isobutane four carbon atoms are joined with one carbon so formula is same but they are structurally different. Now we can take help using the equation, we know that butane is burned with oxygen for its combustion. Equation can be written for it as-
${C_4}{H_{10}}\, + \,\dfrac{{13}}{2}{O_2}\, \to \,4C{O_2}\, + \,5{H_2}O$
As you see that butane reacts with oxygen and gives water and carbon dioxide. Now if one mole reacts to give this carbon dioxide and water. One mole means $58\,g\,of\,{C_4}{H_{10}}$ requires oxygen in amount-
$58\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{13}}{2}\, \times 32$
Here, $32g$ is the mass of oxygen gas, one atom of oxygen is having $16\,g\,mo{l^{ - 1}}$ , so on solving it we get- $58\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,208\,g$ .As $208\,g$ of oxygen is required for reacting with $58\,g\,of\,{C_4}{H_{10}}$ thus now firstly, we find out the amount of oxygen for one gram of butane.
$1\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{208\,}}{{58}}\,g$
For finding the amount of oxygen for reacting with one kilograms is calculated by multiplying the above value by $1000\,g$
$1000\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{208\,}}{{58}}\, \times 1000\,g$ = $3.586\,g$
From the above calculation we find out that the required amount of oxygen is needed for reacting with one kilogram of butane or isobutane.
Hence the correct answer is option ‘C’.
Note: This is a type of basic mathematical calculation. Don’t forget to take molar mass of oxygen as $32g$ instead of $16g$ because there are two atoms of oxygen combined. At last multiply the calculated amount of oxygen with $1000\,g$ for one kilograms. Also the chemical formula of both molecules means butane and isobutane are the same, only there is difference in their representation of formula.
Complete step-by-step answer:
We have formula for butane as ${C_4}{H_{10}}$ and for isobutane the formula is same, the different is that in isobutane four carbon atoms are joined with one carbon so formula is same but they are structurally different. Now we can take help using the equation, we know that butane is burned with oxygen for its combustion. Equation can be written for it as-
${C_4}{H_{10}}\, + \,\dfrac{{13}}{2}{O_2}\, \to \,4C{O_2}\, + \,5{H_2}O$
As you see that butane reacts with oxygen and gives water and carbon dioxide. Now if one mole reacts to give this carbon dioxide and water. One mole means $58\,g\,of\,{C_4}{H_{10}}$ requires oxygen in amount-
$58\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{13}}{2}\, \times 32$
Here, $32g$ is the mass of oxygen gas, one atom of oxygen is having $16\,g\,mo{l^{ - 1}}$ , so on solving it we get- $58\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,208\,g$ .As $208\,g$ of oxygen is required for reacting with $58\,g\,of\,{C_4}{H_{10}}$ thus now firstly, we find out the amount of oxygen for one gram of butane.
$1\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{208\,}}{{58}}\,g$
For finding the amount of oxygen for reacting with one kilograms is calculated by multiplying the above value by $1000\,g$
$1000\,g\,of\,{C_4}{H_{10}}\,requires\,{O_2} = \,\dfrac{{208\,}}{{58}}\, \times 1000\,g$ = $3.586\,g$
From the above calculation we find out that the required amount of oxygen is needed for reacting with one kilogram of butane or isobutane.
Hence the correct answer is option ‘C’.
Note: This is a type of basic mathematical calculation. Don’t forget to take molar mass of oxygen as $32g$ instead of $16g$ because there are two atoms of oxygen combined. At last multiply the calculated amount of oxygen with $1000\,g$ for one kilograms. Also the chemical formula of both molecules means butane and isobutane are the same, only there is difference in their representation of formula.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

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

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

