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To calculate the heat released when 1250 kg of ammonia (NH₃) is formed, follow these steps:
1. Convert the mass of NH₃ to grams:
The given mass of NH₃ is 1250 kg. Since 1 kg = 1000 grams, the mass in grams is:
[tex]\[ \text{mass}_{NH3} = 1250 \, \text{kg} \times 1000 \, (\text{g/kg}) = 1250000 \, \text{g} \][/tex]
2. Calculate the molar mass of NH₃:
The molar mass of NH₃ can be calculated based on the atomic masses of nitrogen (N) and hydrogen (H):
[tex]\[ \text{molar mass}_{NH3} = 14 \, (\text{g/mol}) + 3 \times 1 \, (\text{g/mol}) = 17 \, \text{g/mol} \][/tex]
3. Determine the number of moles of NH₃:
Using the mass and molar mass, the number of moles of NH₃ can be calculated as:
[tex]\[ \text{moles}_{NH3} = \frac{\text{mass}_{NH3}}{\text{molar mass}_{NH3}} = \frac{1250000 \, \text{g}}{17 \, \text{g/mol}} = 73529.41 \, \text{moles} \][/tex]
4. Determine the ΔH for the reaction:
For the given reaction, the enthalpy change (ΔH) is provided as -91.8 kJ per 2 moles of NH₃ formed. The heat released per 1 mole of NH₃ can then be calculated as:
[tex]\[ \Delta H_{1 \, \text{mole}} = \frac{\Delta H}{2} = \frac{-91.8 \, \text{kJ}}{2} = -45.9 \, \text{kJ/mol} \][/tex]
5. Calculate the total heat (q) released:
Finally, the total heat released when 1250 kg of NH₃ is formed is determined by multiplying the heat released per mole of NH₃ by the total number of moles:
[tex]\[ q = \Delta H_{1 \, \text{mole}} \times \text{moles}_{NH3} = -45.9 \, \text{kJ/mol} \times 73529.41 \, \text{moles} = -3375000.0 \, \text{kJ} \][/tex]
Therefore, the heat released when 1250 kg of ammonia is formed is [tex]\( -3375000.0 \, \text{kJ} \)[/tex].
1. Convert the mass of NH₃ to grams:
The given mass of NH₃ is 1250 kg. Since 1 kg = 1000 grams, the mass in grams is:
[tex]\[ \text{mass}_{NH3} = 1250 \, \text{kg} \times 1000 \, (\text{g/kg}) = 1250000 \, \text{g} \][/tex]
2. Calculate the molar mass of NH₃:
The molar mass of NH₃ can be calculated based on the atomic masses of nitrogen (N) and hydrogen (H):
[tex]\[ \text{molar mass}_{NH3} = 14 \, (\text{g/mol}) + 3 \times 1 \, (\text{g/mol}) = 17 \, \text{g/mol} \][/tex]
3. Determine the number of moles of NH₃:
Using the mass and molar mass, the number of moles of NH₃ can be calculated as:
[tex]\[ \text{moles}_{NH3} = \frac{\text{mass}_{NH3}}{\text{molar mass}_{NH3}} = \frac{1250000 \, \text{g}}{17 \, \text{g/mol}} = 73529.41 \, \text{moles} \][/tex]
4. Determine the ΔH for the reaction:
For the given reaction, the enthalpy change (ΔH) is provided as -91.8 kJ per 2 moles of NH₃ formed. The heat released per 1 mole of NH₃ can then be calculated as:
[tex]\[ \Delta H_{1 \, \text{mole}} = \frac{\Delta H}{2} = \frac{-91.8 \, \text{kJ}}{2} = -45.9 \, \text{kJ/mol} \][/tex]
5. Calculate the total heat (q) released:
Finally, the total heat released when 1250 kg of NH₃ is formed is determined by multiplying the heat released per mole of NH₃ by the total number of moles:
[tex]\[ q = \Delta H_{1 \, \text{mole}} \times \text{moles}_{NH3} = -45.9 \, \text{kJ/mol} \times 73529.41 \, \text{moles} = -3375000.0 \, \text{kJ} \][/tex]
Therefore, the heat released when 1250 kg of ammonia is formed is [tex]\( -3375000.0 \, \text{kJ} \)[/tex].
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