Answer :
Final answer:
The total volume of gas produced by the complete decomposition of 1.48 kg of ammonium nitrate, at the provided conditions, is approximately 428 liters.
Explanation:
The question requires us to find the total volume of gas produced by the complete decomposition of a certain amount of ammonium nitrate. This falls under the domain of chemistry, specifically stoichiometry.
Firstly, the balanced equation for the decomposition of ammonium nitrate is NH4NO3 = N2 + O2 + 2H2O. The molar mass of NH4NO3 (ammonium nitrate) is approximately 80 g/mol. This means that 1.48kg (or 1480g) of NH4NO3 corresponds to about 18.5 mol.
From the stoichiometry of the reaction, we know that one mole of gas is produced for every mole of NH4NO3 that reacts. Hence, 18.5 mol of NH4NO3 will produce 18.5 mol of gas. To find the volume this gas occupies, we can use the ideal gas law: PV = nRT. With the given conditions of the problem (T = 116 °C = 389 K; P = 742 mmHg = 0.976 atm), the volume V can be found to be approximately 428 liters.
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