Question

Q3. (20 pts) Ammonium nitrite NH NO,, decomposes according to the following chemical equation NH,NO, (8) ► N, (g) + 2H2O(g) What is the total volume of products obtained when 128 g NH.NO, decomposes at 819 °C and 2.00 atm ? (Atomic weight of N=14.00 g/mol, H=1.01g/mol, O= 16.00 g/mol)

Answer 1

The decomposition reaction of ammonium nitrite is

1 mol ammonium nitrite produces 1 mol nitrogen and 2 mol water in gaseous state.

Given that, mass of
NH4NO2
decomposed = 128 g

Molar mass of
NH4NO2
= 2 × (At. Wt of N) + 4 × (At. Wt of H) + 2 × (At. Wt of O)

Molar mass = 2 × (14) + 4 × (1) + 2 × (16) = 64.00 g/mol

No of moles of
NH4NO2

128(g) 64(g/mol) = 2mol.NH4NO2

We already said that, 1 mol
NH4NO2
produces 1 mol
N2
and 2 mol
H2O
​​​​​​.

Therefore 2 mol
NH4NO2
produce 2 mol
N2
and 4 mol .

Total no of moles of products formed (n) =  6 moles

Also given that, Pressure (P) = 2 atm

Temperature (T) = 819°C = (273 + 819)K = 1092 K

Gas constant, R =  0.08206 atm.L/mol.K

​​​​​​According to Ideal gas equation,

$\small \bold {P*V=n*R*T}$

n *R*T V P

By substituting the above values in the equation,

V 6(mol) * 0.08206( mat.m) * 1092(K) 2 (atm)

Y= 268.8L

The total volume of products formed when 128 g ammonium nitrite decomposed,

V = 268.8 L