Answer
Verified
399.9k+ views
Hint: The general combustion reaction of a hydrocarbon to produce \[C{{O}_{2}}\] and \[{{H}_{2}}O~\] is given by:
\[{{C}_{x}}{{H}_{y}}+(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O\]
Complete step by step solution:
We know that,
A mole of any gas at STP occupies 22.4 litres in volume.
Since, it is initially given that the initial volume of Hydrocarbon is 10 ml.
Therefore, the initial number of moles of hydrocarbon is given as:
$n=\dfrac{0.01}{22.4}\approx 0.45\text{ millimoles}$
We also observe the ratio of volume of hydrocarbon to the eventual volume of the products \[C{{O}_{2}}\] and \[{{H}_{2}}O~\] is:
10:40:50 = 1:4:5
Therefore, the amount of millimoles of \[C{{O}_{2}}\] and \[{{H}_{2}}O~\]respectively is:
$\begin{align}
& {{n}_{C{{O}_{2}}}}=0.45\times 4=1.8\text{ millimoles} \\
& {{n}_{{{H}_{2}}O}}=0.45\times 5=2.23\text{ millimoles} \\
\end{align}$
With this, we observe that one mole of hydrocarbon produces four moles of \[C{{O}_{2}}\],
Therefore, we can safely conclude that the hydrocarbon in question contains 4 atoms of Carbon. This helps us eliminate option b) as a possible answer straight off the bat.
Let us now plug this value of x into the above formula:
\[{{C}_{4}}{{H}_{y}}+(4+\dfrac{y}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O\]
Now let us try and solve for y.
We observe that, one mole of hydrocarbon results in the formation of five moles of water.
That implies, $\dfrac{y}{2}=5$
Thus, we can conclude that the value of y is 10.
Now plugging in the values of both x and y into the reaction of a general hydrocarbon with Oxygen, we observe that:
\[\begin{align}
& {{C}_{4}}{{H}_{10}}+(4+\dfrac{10}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{10}{2}{{H}_{2}}O \\
& {{C}_{4}}{{H}_{10}}+6.5{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O \\
\end{align}\]
Thus, we can effectively conclude that the answer to this question is (D).
Note:
Remember that all hydrocarbons will give only carbon dioxide and hydrogen gas upon combustion, the amount of them will depend on the number of carbon and hydrogen atoms present in the molecule. To solve this question, you will need to relate the volume of gases to their concentration in moles, the only way you can relate them is by using the fact that “all gases have volume of 22.4 L when their concentration is 1 mole at STP” because here the density of gases are not provided.
\[{{C}_{x}}{{H}_{y}}+(x+\dfrac{y}{4}){{O}_{2}}\to xC{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O\]
Complete step by step solution:
We know that,
A mole of any gas at STP occupies 22.4 litres in volume.
Since, it is initially given that the initial volume of Hydrocarbon is 10 ml.
Therefore, the initial number of moles of hydrocarbon is given as:
$n=\dfrac{0.01}{22.4}\approx 0.45\text{ millimoles}$
We also observe the ratio of volume of hydrocarbon to the eventual volume of the products \[C{{O}_{2}}\] and \[{{H}_{2}}O~\] is:
10:40:50 = 1:4:5
Therefore, the amount of millimoles of \[C{{O}_{2}}\] and \[{{H}_{2}}O~\]respectively is:
$\begin{align}
& {{n}_{C{{O}_{2}}}}=0.45\times 4=1.8\text{ millimoles} \\
& {{n}_{{{H}_{2}}O}}=0.45\times 5=2.23\text{ millimoles} \\
\end{align}$
With this, we observe that one mole of hydrocarbon produces four moles of \[C{{O}_{2}}\],
Therefore, we can safely conclude that the hydrocarbon in question contains 4 atoms of Carbon. This helps us eliminate option b) as a possible answer straight off the bat.
Let us now plug this value of x into the above formula:
\[{{C}_{4}}{{H}_{y}}+(4+\dfrac{y}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{y}{2}{{H}_{2}}O\]
Now let us try and solve for y.
We observe that, one mole of hydrocarbon results in the formation of five moles of water.
That implies, $\dfrac{y}{2}=5$
Thus, we can conclude that the value of y is 10.
Now plugging in the values of both x and y into the reaction of a general hydrocarbon with Oxygen, we observe that:
\[\begin{align}
& {{C}_{4}}{{H}_{10}}+(4+\dfrac{10}{4}){{O}_{2}}\to 4C{{O}_{2}}+\dfrac{10}{2}{{H}_{2}}O \\
& {{C}_{4}}{{H}_{10}}+6.5{{O}_{2}}\to 4C{{O}_{2}}+5{{H}_{2}}O \\
\end{align}\]
Thus, we can effectively conclude that the answer to this question is (D).
Note:
Remember that all hydrocarbons will give only carbon dioxide and hydrogen gas upon combustion, the amount of them will depend on the number of carbon and hydrogen atoms present in the molecule. To solve this question, you will need to relate the volume of gases to their concentration in moles, the only way you can relate them is by using the fact that “all gases have volume of 22.4 L when their concentration is 1 mole at STP” because here the density of gases are not provided.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Summary of the poem Where the Mind is Without Fear class 8 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Write an application to the principal requesting five class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE