
An amount of 2 moles \[KCl{{O}_{3}}\] is decomposed completely to produce \[{{O}_{2}}\] gas. How many moles of butane \[{{C}_{4}}{{H}_{8}}\] can be burst completely by the \[{{O}_{2}}\] gas produced?
(A) 0.5
(B) 1.0
(C) 2.0
(D) 3.0
Answer
156.6k+ views
Hint: We should know how many moles of oxygen gas is going to be released by the decomposition of 2 moles of potassium chlorate.
The decomposition of potassium chlorate is as follows.
\[\begin{align}
& KCl{{O}_{3}}\to KCl+\dfrac{3}{2}{{O}_{2}} \\
& 2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}} \\
\end{align}\]
Complete step by step answer:
Decomposition of \[KCl{{O}_{3}}\]is as follows.
\[\begin{align}
& KCl{{O}_{3}}\to KCl+\dfrac{3}{2}{{O}_{2}} \\
& 2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}} \\
\end{align}\]
From the above reaction we can say that 1 mole of \[KCl{{O}_{3}}\]decomposes and forms \[\dfrac{3}{2}\]moles of oxygen and one mole of potassium chloride (KCl).
2 moles of \[KCl{{O}_{3}}\]decomposes and forms 3 moles of oxygen and 2 moles of potassium chloride (KCl) as the products.
The oxygen formed in the above reaction reacts with butane (\[{{C}_{4}}{{H}_{8}}\]).
The burning of butane with oxygen (Combustion) is as follows.
\[{{C}_{4}}{{H}_{8}}+6{{O}_{2}}\to 4C{{O}_{2}}+4{{H}_{2}}O\]
1 mole of Butane reacts with 6 moles of oxygen and forms four moles of carbon dioxide and 4 moles of water.
Means to burst the butane completely there is a need of 6 moles of oxygen.
6 moles of oxygen reacts with one mole of butane.
Therefore 3 moles of oxygen reacts with 0.5 moles of butane.
Then 3 moles of oxygen can burst 0.5 moles of butane completely.
So, the correct option is A.
Note: To burn organic compounds like butane we need oxygen. Burning of hydrocarbons in presence of oxygen produces carbon dioxide and water as the products. Burning of hydrocarbons in presence of oxygen is called combustion.
The decomposition of potassium chlorate is as follows.
\[\begin{align}
& KCl{{O}_{3}}\to KCl+\dfrac{3}{2}{{O}_{2}} \\
& 2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}} \\
\end{align}\]
Complete step by step answer:
Decomposition of \[KCl{{O}_{3}}\]is as follows.
\[\begin{align}
& KCl{{O}_{3}}\to KCl+\dfrac{3}{2}{{O}_{2}} \\
& 2KCl{{O}_{3}}\to 2KCl+3{{O}_{2}} \\
\end{align}\]
From the above reaction we can say that 1 mole of \[KCl{{O}_{3}}\]decomposes and forms \[\dfrac{3}{2}\]moles of oxygen and one mole of potassium chloride (KCl).
2 moles of \[KCl{{O}_{3}}\]decomposes and forms 3 moles of oxygen and 2 moles of potassium chloride (KCl) as the products.
The oxygen formed in the above reaction reacts with butane (\[{{C}_{4}}{{H}_{8}}\]).
The burning of butane with oxygen (Combustion) is as follows.
\[{{C}_{4}}{{H}_{8}}+6{{O}_{2}}\to 4C{{O}_{2}}+4{{H}_{2}}O\]
1 mole of Butane reacts with 6 moles of oxygen and forms four moles of carbon dioxide and 4 moles of water.
Means to burst the butane completely there is a need of 6 moles of oxygen.
6 moles of oxygen reacts with one mole of butane.
Therefore 3 moles of oxygen reacts with 0.5 moles of butane.
Then 3 moles of oxygen can burst 0.5 moles of butane completely.
So, the correct option is A.
Note: To burn organic compounds like butane we need oxygen. Burning of hydrocarbons in presence of oxygen produces carbon dioxide and water as the products. Burning of hydrocarbons in presence of oxygen is called combustion.
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