
If a bottle of nail polish remover contains $176\,ml$ of acetone, how much heat would be released by its complete combustion? The density of acetone is $0.788gm{{l}^{-1}}$
Answer
558.9k+ views
Hint:Combustion is a process when a substance reacts with oxygen to form a product and release heat. Combustion of hydrocarbons in air results in an exothermic process. There are two types of combustion such as complete and incomplete.
Complete step by step solution:
The reaction for complete combustion can be written as:
$C{{H}_{3}}COC{{H}_{3(l)}}+4{{O}_{2(g)}}\to 3C{{O}_{2(g)}}+3{{H}_{2}}{{O}_{(l)}}+heat$
Here, it is given that the volume of acetone is $176\,ml$
The density of acetone is $0.788gm{{l}^{-1}}$
To calculate the mass of the acetone,
$\rho =\dfrac{W}{v}$
where, $\rho $ is the density, $W$ is the mass and $v$ is the volume.
$\Rightarrow W=\rho \times v$
Now, substituting the values in the above formula, we get,
$\Rightarrow W=\rho \times v$
$\Rightarrow W=0.788\times 176$
$\Rightarrow W=138.688\,g$
Now we have to calculate the moles of acetone.
Molar mass of acetone is $58.08\,g$
$n=\dfrac{W}{M}$
where, $n$ is the number of moles, $W$ is weight of acetone and $M$ is the molar mass of acetone.
Now, substituting the values in the above formula, we get,
$n=\dfrac{138.688}{58.08}$
$n=2.39\,mol$
Molar enthalpy of combustion of acetone is $-1790\,kJ\,mo{{l}^{-1}}$
Molar enthalpy of combustion is used to calculate the amount of heat released when an amount of substance is present.
We take the product of moles of acetone and molar enthalpy of combustion
$q=n\times \Delta {{H}_{C}}$
where, $q$ is the heat released, $n$ is the number of moles of acetone and $\Delta {{H}_{C}}$ is the molar enthalpy of combustion.
Now, substituting the above values in the formula, we get
$q=2.39\times -1790$
$\Rightarrow q=-4278.1\,kJ$
Therefore, the heat released by complete combustion of acetone is $-4278.1\,kJ$.
Note:Molar enthalpy of combustion is defined as a substance that releases heat when one mole of substance undergoes combustion at constant pressureWhen hydrocarbons undergo complete combustion in presence of excess oxygen the products formed are carbon dioxide and water.
$Hydrocarbon+excess\,{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$
Complete step by step solution:
The reaction for complete combustion can be written as:
$C{{H}_{3}}COC{{H}_{3(l)}}+4{{O}_{2(g)}}\to 3C{{O}_{2(g)}}+3{{H}_{2}}{{O}_{(l)}}+heat$
Here, it is given that the volume of acetone is $176\,ml$
The density of acetone is $0.788gm{{l}^{-1}}$
To calculate the mass of the acetone,
$\rho =\dfrac{W}{v}$
where, $\rho $ is the density, $W$ is the mass and $v$ is the volume.
$\Rightarrow W=\rho \times v$
Now, substituting the values in the above formula, we get,
$\Rightarrow W=\rho \times v$
$\Rightarrow W=0.788\times 176$
$\Rightarrow W=138.688\,g$
Now we have to calculate the moles of acetone.
Molar mass of acetone is $58.08\,g$
$n=\dfrac{W}{M}$
where, $n$ is the number of moles, $W$ is weight of acetone and $M$ is the molar mass of acetone.
Now, substituting the values in the above formula, we get,
$n=\dfrac{138.688}{58.08}$
$n=2.39\,mol$
Molar enthalpy of combustion of acetone is $-1790\,kJ\,mo{{l}^{-1}}$
Molar enthalpy of combustion is used to calculate the amount of heat released when an amount of substance is present.
We take the product of moles of acetone and molar enthalpy of combustion
$q=n\times \Delta {{H}_{C}}$
where, $q$ is the heat released, $n$ is the number of moles of acetone and $\Delta {{H}_{C}}$ is the molar enthalpy of combustion.
Now, substituting the above values in the formula, we get
$q=2.39\times -1790$
$\Rightarrow q=-4278.1\,kJ$
Therefore, the heat released by complete combustion of acetone is $-4278.1\,kJ$.
Note:Molar enthalpy of combustion is defined as a substance that releases heat when one mole of substance undergoes combustion at constant pressureWhen hydrocarbons undergo complete combustion in presence of excess oxygen the products formed are carbon dioxide and water.
$Hydrocarbon+excess\,{{O}_{2}}\to C{{O}_{2}}+{{H}_{2}}O$
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