Ammonium nitrate, NH4NO3, can decompose explosively when heated to a high temperature according to the equation 2 NH4NO3(s) ® 2 N2(g) + 4 H2O(g) + O2(g) If 100.0 g of ammonium nitrate decomposes at 450 oC, how many liters of gaseous products would be formed? (Assume atmospheric pressure is 1.00 atm) ### Decomposition of Ammonium NitrateAmmonium nitrate, \(\text{NH}_4\text{NO}_3\), can decompose explosively when heated to a high temperature according to the following equation:\[ 2 \, \text{NH}_4\text{NO}_3(s) \rightarrow 2 \, \text{N}_2(g) + 4 \, \text{H}_2\text{O}(g) + \text{O}_2(g) \]### Problem StatementIf 100.0 g of ammonium nitrate decomposes at 450 °C, how many liters of gaseous products would be formed? (Assume atmospheric pressure is 1.00 atm)### Explanation of Chemical Reaction- **Reactants**: - Ammonium nitrate (\(\text{NH}_4\text{NO}_3\)) in solid form. - **Products**: - Nitrogen gas (\(\text{N}_2\)) - Water vapor (\(\text{H}_2\text{O}\)) - Oxygen gas (\(\text{O}_2\)) The equation shows that decomposition of solid ammonium nitrate results in the formation of multiple gases.### Calculation Guide- **Molar Masses**: - \(\text{NH}_4\text{NO}_3\) = 80.04 g/mol- **Using the Ideal Gas Law**: - Calculate moles of \(\text{NH}_4\text{NO}_3\) - Use stoichiometry to find moles of gaseous products - Apply Ideal Gas Law to find the volume of gases produced at given conditions (450 °C, 1.00 atm)This explanation will assist in understanding the reaction dynamics and calculating the volume of gas generated from the decomposition of ammonium nitrate at high temperatures.

Interpretation- To determine the how many liters of gaseous products would be formed. When NH4NO3, can decompose explosively when heated to a high temperature according to the equation- 2 NH4NO3(s) → 2 N2(g) + 4 H2O(g) + O2(g) and given 100.0 g of ammonium nitrate decomposes at 450 oC and assume pressure is 1 atm . Concept- To determine the volume of gaseous product ( or liters of gaseous products ) we use ideal gas equation , according to which - PV= nRT where- P - pressure of gas V - volume n - number of mol R- gas constant T - Temperature Given that -100 g of Ammonium nitrate decompose at 450°CSo- Weight of Ammonium Nitrate (w) = 100gTemperature (T) = 450°C T = (450+273)K T = 723 KAssume Pressure (P) = 1 atm Volume of gaseous product formed by NH4NO3(s) =by PV=nRT → ideal gas equation∵ number of moles(n) = w (mass)Mw(molar mass)So PV = wNH4NO3RTMNH4NO3put the value of P,T.wNH4NO3 and Molar mass (MNH4NO3) & gas constant R~PV = wNH4NO3RTMNH4NO31atm x V = 100g x 0.0821 L-atm K-1mol-1 x 723 K80.043 g mol-1V(in L) = 100 g x 0.0821 L - atm K-1x mol-1x 723 K1 atm x 80.043 g mol-1Volume in ( liter) = 100 x 0.0821 x 723 L1 x 80.043 So Volume = 5935.83 L80.043Volume of gaseous product = 74.15 liter Ans.