Answer
Verified
452.1k+ views
Hint:
The given sample is a mixture of gases which consist of \[{C_4}{H_{10}}\], \[C{H_4}\] and CO. Now each of these gases contribute to the physical properties of the mixture. They do not contribute to the chemical properties as a whole, because they are still unreactive with each other and form no by products.
Complete step by step answer:
The given gaseous mixture is made up of 3 gases namely, \[{C_4}{H_{10}}\], \[C{H_4}\] and CO. Now, the total volume of the gaseous mixture is said to be 200 ml. Out of this, constitutes about 40% of the total volume of the given gaseous mixture. Hence, to calculate the actual volume of \[{C_4}{H_{10}}\]:
Volume of \[{C_4}{H_{10}}\] = 40 % of the entire gaseous mixture
= \[\dfrac{{40}}{{100}} \times \]volume of the entire gaseous mixture
= 0.4 \[ \times \]200ml
= 80 ml
Now the remainder of the gaseous mixture has a volume of about (200-80) = 120 ml.
The remainder of the gaseous mixture includes the gases \[C{H_4}\] and CO. For the ease of calculation, let us consider the volume of \[C{H_4}\] to be ‘x’ ml. Then the volume of CO will be of whatever remains, i.e ‘120 - x’ ml.
The conditions of the reaction of this gaseous mixture is given that the entire mixture is burnt in the presence of excess of oxygen. Considering that one mole of each constituent molecule is used, the reactions of the constituent gases of the mixture with oxygen can be given by:
\[\mathop {C{H_4}}\limits_{'x'ml} + 2{O_2} \to \mathop {C{0_2}}\limits_{'x'ml} + 2{H_2}O\]
\[\mathop {CO}\limits_{(120 - x)ml} + \dfrac{1}{2}{O_2} \to \mathop {C{O_2}}\limits_{(120 - x)ml} \]
\[\mathop {{C_4}{H_{10}}}\limits_{80ml} + \dfrac{{13}}{2}{O_2} \to \mathop {4C{O_2}}\limits_{4 \times 80 = 320ml} + 5{H_2}O\]
Hence the total volume of carbon dioxide produced can be calculated as:
Total volume of \[C{O_2}\]= (x) + (120-x) + (320) = 120 + 320 = 440 ml
Hence, Option C is the correct option.
Note:
The excess oxygen that has been used in the burning the mixture is responsible for deriving carbon dioxide from each of the constituent gases. In case if the oxygen was supplied in a limited quantity, then all the constituents would not have been fully ignited and hence, the amount of carbon dioxide would be less.
The given sample is a mixture of gases which consist of \[{C_4}{H_{10}}\], \[C{H_4}\] and CO. Now each of these gases contribute to the physical properties of the mixture. They do not contribute to the chemical properties as a whole, because they are still unreactive with each other and form no by products.
Complete step by step answer:
The given gaseous mixture is made up of 3 gases namely, \[{C_4}{H_{10}}\], \[C{H_4}\] and CO. Now, the total volume of the gaseous mixture is said to be 200 ml. Out of this, constitutes about 40% of the total volume of the given gaseous mixture. Hence, to calculate the actual volume of \[{C_4}{H_{10}}\]:
Volume of \[{C_4}{H_{10}}\] = 40 % of the entire gaseous mixture
= \[\dfrac{{40}}{{100}} \times \]volume of the entire gaseous mixture
= 0.4 \[ \times \]200ml
= 80 ml
Now the remainder of the gaseous mixture has a volume of about (200-80) = 120 ml.
The remainder of the gaseous mixture includes the gases \[C{H_4}\] and CO. For the ease of calculation, let us consider the volume of \[C{H_4}\] to be ‘x’ ml. Then the volume of CO will be of whatever remains, i.e ‘120 - x’ ml.
The conditions of the reaction of this gaseous mixture is given that the entire mixture is burnt in the presence of excess of oxygen. Considering that one mole of each constituent molecule is used, the reactions of the constituent gases of the mixture with oxygen can be given by:
\[\mathop {C{H_4}}\limits_{'x'ml} + 2{O_2} \to \mathop {C{0_2}}\limits_{'x'ml} + 2{H_2}O\]
\[\mathop {CO}\limits_{(120 - x)ml} + \dfrac{1}{2}{O_2} \to \mathop {C{O_2}}\limits_{(120 - x)ml} \]
\[\mathop {{C_4}{H_{10}}}\limits_{80ml} + \dfrac{{13}}{2}{O_2} \to \mathop {4C{O_2}}\limits_{4 \times 80 = 320ml} + 5{H_2}O\]
Hence the total volume of carbon dioxide produced can be calculated as:
Total volume of \[C{O_2}\]= (x) + (120-x) + (320) = 120 + 320 = 440 ml
Hence, Option C is the correct option.
Note:
The excess oxygen that has been used in the burning the mixture is responsible for deriving carbon dioxide from each of the constituent gases. In case if the oxygen was supplied in a limited quantity, then all the constituents would not have been fully ignited and hence, the amount of carbon dioxide would be less.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE