Question
Question: What total volume, in litre at \({627^0}C\) and \(0.821atm\) could be formed by the decomposition of...
What total volume, in litre at 6270C and 0.821atm could be formed by the decomposition of 16g of NH4NO3 ? Reaction: 2NH4NO3→2N2+O2+4H2O(g)
Solution
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 6270C and 0.821atm that will be formed by the decomposition of NH4NO3.
In the question, it is given that 16gm of NH4NO3 is decomposed. We need to find the number of moles;
Molar mass of NH4NO3 = 80.043g(mol)−1
Moles=mol.massmass(given)
Moles=80.04316
= 0.2 moles
From the balanced reaction equation, we can say that;
2 moles of NH4NO3 → 7 moles of gaseous products
1 mole of NH4NO3 → 27 moles of gaseous products
0.2 moles of NH4NO3 → 27×0.2 moles of gaseous product
→ 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)−1K−1
T = temperature
= 6270C (we need to convert it to kelvin)
= 900K
Now, putting the values;
V=PnRT
V=0.821atm0.7mol×0.0821atmL(mol)−1K−1×900K
V=63L
The volume formed by the decomposition of 16g of NH4NO3 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.