Question

Solid ammonium nitrite, $ N{H_4}N{O_2} $ of $ 320g $ decomposes when heated according to the balanced equation: $ N{H_4}N{O_2} \to {N_2} + 2{H_2}O $ . What total volume of gases at $ 819K $ is emitted by this reaction?

Answer 1

Hint: The weight of solid ammonium nitrite was given, from the weight and molar mass the number of moles will be calculated. From the number of moles and balanced chemical equations, the volume occupied and temperature at STP can be determined, the temperature of gases given and the calculated volume and temperature at STP the volume of gases at given temperature will be calculated.
 $ \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} $
 $ {V_1} $ is volume occupied at STP
 $ {T_1} $ is temperature at STP
 $ {V_2} $ is volume at given temperature
 $ {T_2} $ is given temperature.

Complete Step By Step Answer:
The balanced chemical equation of the decomposition of solid ammonium nitrite is $ N{H_4}N{O_2} \to {N_2} + 2{H_2}O $
From the above equation, it was clear that one mole of ammonium nitrite produces $ 3 $ moles of gases.
The weight of ammonium nitrite is given as $ 320g $
The molar mass of ammonium nitrite is $ 64gmo{l^{ - 1}} $
Thus, the number of moles will be $ \dfrac{{320}}{{64}} = 5moles $
One mole of ammonium nitrite produces $ 3 $ moles of gases. Thus, $ 5 $ moles of ammonium nitrite produces $ 3 \times 5 = 15moles $ of gases.
One mole occupies $ 22.4L $ at $ 273K $ and $ 1atm $ which is at STP.
Thus, $ 15moles $ of produced gases occupy $ 22.4 \times 15 = 336L $
The given temperature is $ 819K $
Substitute these values in the above formula
 $ {V_2} = 336 \times \dfrac{{819}}{{273}} = 1008L $
Thus, $ 1008L $ volume of gases at $ 819K $ is emitted by the given reaction.

Note:
While calculating the number of moles, the exact molar mass of ammonium nitrite must be taken. The volume occupied at STP can be calculated from the total moles of produced gas, where the pressure is constant which is $ 1atm $ , only temperature and pressure were changed.