A mixture of ${C_3}{H_8}(g)\& {O_2}$having total volume $100ml$ in an Eudiometer tube is sparked & it is observed that a contraction of $45ml$ is observed what can be the composition of reacting mixture.
A) $15\;ml\;{C_3}{H_8}\;\& \;85ml\;{O_2}$
B) $25\;ml\;{C_3}{H_8}\;\& \;75ml\;{O_2}$
C) $45\;ml\;{C_3}{H_8}\;\& \;55ml\;{O_2}$
D) $55\;ml\;{C_3}{H_8}\;\& \;45ml\;{O_2}$
Answer
582.9k+ views
Hint:We know that propane reacts with oxygen to produce carbon dioxide and water molecules. And we also know that moles of a substance is equivalent to the volume of that substance at standard conditions of temperature and pressure.
Complete solution:
As we know that one mole of propane reacts with five moles of oxygen and results in the formation of three carbon dioxide and four water molecules. We also know that the number of moles is equivalent to the ratio of volume of the gas to $22.4L$.
Thus, we can also say that $1ml$ of propane reacts with $5ml$ of oxygen to give $3ml$ of carbon dioxide and $4ml$ of water.
So, let us assume that propane acts as a limiting reagent which means that it will be completely consumed. So, when $15ml$ of propane reacts with $75ml$ of oxygen gas, $45ml$ of carbon dioxide will be obtained.
And we are given that the contraction is observed as $45ml$. Hence, the contraction in the volume will be equivalent to the sum of all the volumes so it is given as: $15 + 75 - 45 = 45ml$.
Similarly, if we assume oxygen as the limiting reagent. So when $25ml$ propane reacts with $75ml$ of carbon dioxide will be formed, then the contraction will be given as: $5 + 25 - 75 = 45ml$.
Therefore, we can assume that this contraction is possible when the reaction mixture can be either $15\;ml\;{C_3}{H_8}\;\& \;85ml\;{O_2}$ or $25\;ml\;{C_3}{H_8}\;\& \;75ml\;{O_2}$.
Hence, from the above explanation we can say that the correct answers are (A) and (B).
Note:Always remember that the limiting reagent is that reactant which will be completely consumed first during the course of a reaction. Also remember that the contraction in the volume is basically the term which is defined as the decrease in the volume of that particular substance.
Complete solution:
As we know that one mole of propane reacts with five moles of oxygen and results in the formation of three carbon dioxide and four water molecules. We also know that the number of moles is equivalent to the ratio of volume of the gas to $22.4L$.
Thus, we can also say that $1ml$ of propane reacts with $5ml$ of oxygen to give $3ml$ of carbon dioxide and $4ml$ of water.
So, let us assume that propane acts as a limiting reagent which means that it will be completely consumed. So, when $15ml$ of propane reacts with $75ml$ of oxygen gas, $45ml$ of carbon dioxide will be obtained.
And we are given that the contraction is observed as $45ml$. Hence, the contraction in the volume will be equivalent to the sum of all the volumes so it is given as: $15 + 75 - 45 = 45ml$.
Similarly, if we assume oxygen as the limiting reagent. So when $25ml$ propane reacts with $75ml$ of carbon dioxide will be formed, then the contraction will be given as: $5 + 25 - 75 = 45ml$.
Therefore, we can assume that this contraction is possible when the reaction mixture can be either $15\;ml\;{C_3}{H_8}\;\& \;85ml\;{O_2}$ or $25\;ml\;{C_3}{H_8}\;\& \;75ml\;{O_2}$.
Hence, from the above explanation we can say that the correct answers are (A) and (B).
Note:Always remember that the limiting reagent is that reactant which will be completely consumed first during the course of a reaction. Also remember that the contraction in the volume is basically the term which is defined as the decrease in the volume of that particular substance.
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