Question

Определите сколько аммиака (мл) выделится при разложении 0,5г карбоната аммония при 250 С

Answer 1

(NH4)2CO3 = CO2 + H2O + 2NH3

n((NH4)2CO3) = m((NH4)2CO3)/M((NH4)2CO3) = 0,5/96 = 0,005 моль

n(NH3) = 2*n((NH4)CO3) = 2*0,005= 0,01 моль

V(NH3) = n(NH3)*Vm = 0,01*22,4 = 0,224 л  = 224 мл - это объем при нормальных условиях.

Так как давления при 250 градусах не дано, то и объем аммиака при этой температуре вычислить не можем (или задача не дописана).

А вообще, для разложения карбоната аммония по такому уравнению не нужна температура в 250 градусов, достаточно 60.

Answer 2

Calculating Ammonia Production from Ammonium Carbonate Decomposition

To calculate the amount of ammonia produced from the decomposition of 0.5g of ammonium carbonate at 250°C, we can use the chemical equation for the decomposition of ammonium carbonate:

NH4HCO3 → NH3 + CO2 + H2O

According to the stoichiometry of the reaction, 1 mole of ammonium carbonate decomposes to produce 1 mole of ammonia. We can use this information to calculate the amount of ammonia produced.

Calculation

1. Molar Mass of Ammonium Carbonate (NH4HCO3) - The molar mass of NH4HCO3 is 79.056 g/mol.

2. Moles of Ammonium Carbonate - Moles = Mass / Molar Mass - Moles = 0.5g / 79.056 g/mol - Moles ≈ 0.00632 mol

3. Moles of Ammonia Produced - Since the stoichiometry of the reaction is 1:1, the moles of ammonia produced will be the same as the moles of ammonium carbonate decomposed. - Moles of Ammonia = 0.00632 mol

4. Volume of Ammonia Produced at Standard Temperature and Pressure (STP) - 1 mole of any gas occupies 22.4 liters at STP. - Volume of Ammonia = Moles of Ammonia * 22.4 L/mol - Volume of Ammonia ≈ 0.1416 L

Conclusion

Approximately 0.1416 liters of ammonia will be produced from the decomposition of 0.5g of ammonium carbonate at 250°C.