
In the formula ${{H}_{2}}{{O}_{2}}\to {{H}_{2}}O+{{O}_{2}}$ , how many grams of ${{O}_{2}}$ are produced from the decomposition of $68\,g$ of ${{H}_{2}}{{O}_{2}}$ ?
Answer
541.2k+ views
Hint: Mole concept gives the relationship between the number of moles, weight and molar mass of the compound. Molar mass is calculated by adding up the atomic masses of the element combined to form a molecule.
Formula used:$n=\dfrac{w}{m}$
where, $n$ is the number of moles, $W$ is the weight of the compound and $M$ is the molar mass of the compound.
Complete step-by-step answer:The given equation is not balanced; therefore, we need to balance this equation first.
$2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}$
Here, it is given that the weight of hydrogen peroxide $({{H}_{2}}{{O}_{2}})$ is $68\,g$
The molar mass of hydrogen peroxide is
${{m}_{{{H}_{2}}{{O}_{2}}}}=2\times 1+2\times 16$
${{m}_{{{H}_{2}}{{O}_{2}}}}=34$
Now, we will calculate the moles of hydrogen peroxide using the values given,
$n=\dfrac{w}{m}$
Now, substituting the values in the above equation, we get,
$\Rightarrow n=\dfrac{68}{34}$
$\Rightarrow n=2\,moles$
After that, we will compare the moles of hydrogen peroxide and oxygen gas according to the balanced equation.
The mole ratio between ${{H}_{2}}{{O}_{2}}$ and ${{O}_{2}}$ gas is
$\Rightarrow (mol){{H}_{2}}{{O}_{2}}:(mol){{O}_{2}}$
$\Rightarrow 2:1$
This means that the moles of oxygen produced in this reaction will be exactly half of the moles of hydrogen peroxide reacting.
$mol({{O}_{2}})=\dfrac{mol({{H}_{2}}{{O}_{2}})}{2}$
Now, we will substitute the value of moles of ${{H}_{2}}{{O}_{2}}$ that we have calculated
$mol({{O}_{2}})=\dfrac{2}{2}$
$mol({{O}_{2}})=1\,mol$
Now, we will convert the mole into the weight of oxygen by using the mole concept in the formula.
$n=\dfrac{w}{m}$
where, $n$ is the number of moles, $W$ is the weight of the compound and $M$ is the molar mass of the compound.
Here, $n=1$ and molar mass of ${{O}_{2}}=32\,g\,mo{{l}^{-1}}$
$w=n\times m$
Now, substituting the values in the above formula, we get,
$\Rightarrow w=1\times 32$
$\Rightarrow w=32\,g$
Hence, the weight of oxygen produced from the decomposition of$68\,g$ of ${{H}_{2}}{{O}_{2}}$is $32\,g$
Additional information:Moles is defined as a scientific unit which is used to measure large quantities of atoms and molecules. It is defined as the amount of substance present in a sample.
$1\,mole=6.022\times {{10}^{23}}particles$
This is also called Avogadro's number $({{N}_{A}})$ .
Molar mass is defined as the addition of atomic mass of atoms, combined in a molecule. Here, the molar mass of \[{{H}_{2}}{{O}_{2}}\] is $34g$ .
Moles are defined as the weight of the compound per molar mass of the compound.
Formula mass is different from molecular mass. In formula mass, there is an addition of atomic masses of ions present in an ionic compound, whereas molecular mass is the addition of atomic masses of the atoms in a molecule.
Note:
The molar mass is calculated in $g\,mo{{l}^{-1}}$ .
You should know how to calculate molar mass of a compound and then substitute the values in the mole concept formula.
Formula used:$n=\dfrac{w}{m}$
where, $n$ is the number of moles, $W$ is the weight of the compound and $M$ is the molar mass of the compound.
Complete step-by-step answer:The given equation is not balanced; therefore, we need to balance this equation first.
$2{{H}_{2}}{{O}_{2}}\to 2{{H}_{2}}O+{{O}_{2}}$
Here, it is given that the weight of hydrogen peroxide $({{H}_{2}}{{O}_{2}})$ is $68\,g$
The molar mass of hydrogen peroxide is
${{m}_{{{H}_{2}}{{O}_{2}}}}=2\times 1+2\times 16$
${{m}_{{{H}_{2}}{{O}_{2}}}}=34$
Now, we will calculate the moles of hydrogen peroxide using the values given,
$n=\dfrac{w}{m}$
Now, substituting the values in the above equation, we get,
$\Rightarrow n=\dfrac{68}{34}$
$\Rightarrow n=2\,moles$
After that, we will compare the moles of hydrogen peroxide and oxygen gas according to the balanced equation.
The mole ratio between ${{H}_{2}}{{O}_{2}}$ and ${{O}_{2}}$ gas is
$\Rightarrow (mol){{H}_{2}}{{O}_{2}}:(mol){{O}_{2}}$
$\Rightarrow 2:1$
This means that the moles of oxygen produced in this reaction will be exactly half of the moles of hydrogen peroxide reacting.
$mol({{O}_{2}})=\dfrac{mol({{H}_{2}}{{O}_{2}})}{2}$
Now, we will substitute the value of moles of ${{H}_{2}}{{O}_{2}}$ that we have calculated
$mol({{O}_{2}})=\dfrac{2}{2}$
$mol({{O}_{2}})=1\,mol$
Now, we will convert the mole into the weight of oxygen by using the mole concept in the formula.
$n=\dfrac{w}{m}$
where, $n$ is the number of moles, $W$ is the weight of the compound and $M$ is the molar mass of the compound.
Here, $n=1$ and molar mass of ${{O}_{2}}=32\,g\,mo{{l}^{-1}}$
$w=n\times m$
Now, substituting the values in the above formula, we get,
$\Rightarrow w=1\times 32$
$\Rightarrow w=32\,g$
Hence, the weight of oxygen produced from the decomposition of$68\,g$ of ${{H}_{2}}{{O}_{2}}$is $32\,g$
Additional information:Moles is defined as a scientific unit which is used to measure large quantities of atoms and molecules. It is defined as the amount of substance present in a sample.
$1\,mole=6.022\times {{10}^{23}}particles$
This is also called Avogadro's number $({{N}_{A}})$ .
Molar mass is defined as the addition of atomic mass of atoms, combined in a molecule. Here, the molar mass of \[{{H}_{2}}{{O}_{2}}\] is $34g$ .
Moles are defined as the weight of the compound per molar mass of the compound.
Formula mass is different from molecular mass. In formula mass, there is an addition of atomic masses of ions present in an ionic compound, whereas molecular mass is the addition of atomic masses of the atoms in a molecule.
Note:
The molar mass is calculated in $g\,mo{{l}^{-1}}$ .
You should know how to calculate molar mass of a compound and then substitute the values in the mole concept formula.
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