
What is the mass of $0.5$ moles of $CO_2$?
Answer
516.9k+ views
Hint: As we know that the moles are one of the main units in chemistry. The moles of the molecule are depending on the mass of the molecule and molecular mass of the molecule. Chemical reactions are measured by moles only. The number of equivalents of the reactant is also dependent on the moles of the molecule. The number of moles of the reactant and product are equal in the equilibrium reaction.
Formula used: We must remember that the moles defined as the given mass of the molecule is divided by the molecular mass of the molecule.
\[{\text{Moles}} = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}\]
The molecular weight of the molecule is depending on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atoms in the molecule.
${\text{Molecular weight}} = {\text{Number of the atom}} \times {\text{Atomic weight of the atom}}$
Complete step by step answer:
In carbon dioxide, there is one carbon atom and two oxygen atoms.
The atomic weight of the carbon atom is $12$ .
The atomic weight of the oxygen atom is $16$ .
Now we can calculate the molecular weight of the carbon dioxide as,
$ = 12 \times 1 \times 16 \times 2$
On simplification we get,
$ = 44g$
The molecular weight of the carbon dioxide is $44g$ .
Calculate the mass of $0.5$ moles of $C{O_2}$
\[{\text{Moles}} = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}\]
\[0.5 = \dfrac{{{\text{Mass of }}C{O_2}}}{{44}}\]
We replace the formula to calculate the mass of the carbon dioxide.
\[{\text{Mass of }}C{O_2} = 0.5 \times 44\]
On multiplication we get,
$ = 22g$
The mass of $0.5$ moles of $CO_2$ is $22 g$ .
Note: The number of moles of the reaction if calculated, in this way we can predict the reaction is completed, are not. The number of moles are used to find the yield percentage of the reaction. The moles are also used to find the formation of side products in the chemical reaction. Carbon dioxide is one of the gases used for cooling the reaction. If one reaction carbon dioxide evolves means that reaction is considered as an endothermic reaction.
Formula used: We must remember that the moles defined as the given mass of the molecule is divided by the molecular mass of the molecule.
\[{\text{Moles}} = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}\]
The molecular weight of the molecule is depending on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atoms in the molecule.
${\text{Molecular weight}} = {\text{Number of the atom}} \times {\text{Atomic weight of the atom}}$
Complete step by step answer:
In carbon dioxide, there is one carbon atom and two oxygen atoms.
The atomic weight of the carbon atom is $12$ .
The atomic weight of the oxygen atom is $16$ .
Now we can calculate the molecular weight of the carbon dioxide as,
$ = 12 \times 1 \times 16 \times 2$
On simplification we get,
$ = 44g$
The molecular weight of the carbon dioxide is $44g$ .
Calculate the mass of $0.5$ moles of $C{O_2}$
\[{\text{Moles}} = \dfrac{{{\text{Mass of the molecule}}}}{{{\text{Molecular weight of the molecule}}}}\]
\[0.5 = \dfrac{{{\text{Mass of }}C{O_2}}}{{44}}\]
We replace the formula to calculate the mass of the carbon dioxide.
\[{\text{Mass of }}C{O_2} = 0.5 \times 44\]
On multiplication we get,
$ = 22g$
The mass of $0.5$ moles of $CO_2$ is $22 g$ .
Note: The number of moles of the reaction if calculated, in this way we can predict the reaction is completed, are not. The number of moles are used to find the yield percentage of the reaction. The moles are also used to find the formation of side products in the chemical reaction. Carbon dioxide is one of the gases used for cooling the reaction. If one reaction carbon dioxide evolves means that reaction is considered as an endothermic reaction.
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