Answer
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Hint: The equation which gives the simultaneous effect of pressure and temperature on the volume of a gas is known as ideal gas or equation of state for an ideal gas. We can use the ideal gas equation in this question in order to find the total volume.
Complete answer:
We have to find the total volume at ${627^0}C$ and $0.821atm$ that will be formed by the decomposition of $N{H_4}N{O_3}$.
In the question, it is given that 16gm of $N{H_4}N{O_3}$ is decomposed. We need to find the number of moles;
Molar mass of $N{H_4}N{O_3}$ = $80.043g{(mol)^{ - 1}}$
$Moles = \dfrac{{mass(given)}}{{mol.mass}}$
$Moles = \dfrac{{16}}{{80.043}}$
= $0.2$ moles
From the balanced reaction equation, we can say that;
$2$ moles of $N{H_4}N{O_3}$ $ \to $ $7$ moles of gaseous products
$1$ mole of $N{H_4}N{O_3}$ $ \to $ $\dfrac{7}{2}$ moles of gaseous products
$0.2$ moles of $N{H_4}N{O_3}$ $ \to $ $\dfrac{{7 \times 0.2}}{2}$ moles of gaseous product
$ \to $ $0.7$ moles
Now we can use ideal gas equation to find the volume;
$PV = nRT$
Where, P = pressure
= $0.821atm$ (given)
n= number of moles of gaseous products formed
= $0.7$ moles
R = gas constant
= $0.0821atmL{(mol)^{ - 1}}{K^{ - 1}}$
T = temperature
= ${627^0}C$ (we need to convert it to kelvin)
= $900K$
Now, putting the values;
$V = \dfrac{{nRT}}{P}$
$V = \dfrac{{0.7mol \times 0.0821atmL{{(mol)}^{ - 1}}{K^{ - 1}} \times 900K}}{{0.821atm}}$
$V = 63L$
The volume formed by the decomposition of $16g$ of $N{H_4}N{O_3}$ was found to be $63L$
Additional Information: Ammonium nitrate is the ammonium salt of nitric acid. It has a role as a fertilizer, an explosive and an oxidizing agent. It is an inorganic molecular entity, an ammonium salt and an inorganic nitrate salt.
Note:
In such types of questions before equating stoichiometry we should make sure that the reaction equation is balanced. We made use of the ideal gas equation to find the volume of gaseous products formed.
Complete answer:
We have to find the total volume at ${627^0}C$ and $0.821atm$ that will be formed by the decomposition of $N{H_4}N{O_3}$.
In the question, it is given that 16gm of $N{H_4}N{O_3}$ is decomposed. We need to find the number of moles;
Molar mass of $N{H_4}N{O_3}$ = $80.043g{(mol)^{ - 1}}$
$Moles = \dfrac{{mass(given)}}{{mol.mass}}$
$Moles = \dfrac{{16}}{{80.043}}$
= $0.2$ moles
From the balanced reaction equation, we can say that;
$2$ moles of $N{H_4}N{O_3}$ $ \to $ $7$ moles of gaseous products
$1$ mole of $N{H_4}N{O_3}$ $ \to $ $\dfrac{7}{2}$ moles of gaseous products
$0.2$ moles of $N{H_4}N{O_3}$ $ \to $ $\dfrac{{7 \times 0.2}}{2}$ moles of gaseous product
$ \to $ $0.7$ moles
Now we can use ideal gas equation to find the volume;
$PV = nRT$
Where, P = pressure
= $0.821atm$ (given)
n= number of moles of gaseous products formed
= $0.7$ moles
R = gas constant
= $0.0821atmL{(mol)^{ - 1}}{K^{ - 1}}$
T = temperature
= ${627^0}C$ (we need to convert it to kelvin)
= $900K$
Now, putting the values;
$V = \dfrac{{nRT}}{P}$
$V = \dfrac{{0.7mol \times 0.0821atmL{{(mol)}^{ - 1}}{K^{ - 1}} \times 900K}}{{0.821atm}}$
$V = 63L$
The volume formed by the decomposition of $16g$ of $N{H_4}N{O_3}$ was found to be $63L$
Additional Information: Ammonium nitrate is the ammonium salt of nitric acid. It has a role as a fertilizer, an explosive and an oxidizing agent. It is an inorganic molecular entity, an ammonium salt and an inorganic nitrate salt.
Note:
In such types of questions before equating stoichiometry we should make sure that the reaction equation is balanced. We made use of the ideal gas equation to find the volume of gaseous products formed.
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