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Ammonium nitrate can decompose explosively when heated according to the equation:

2 NH₄NO₃(s) → 2 N₂(g) +4 H₂O (g) + O₂ (g)
How many LITERS OF GAS would be formed at 450degrees C and 1.00 atm pressure by explosion of 450 g of NH₄NO₃?

User Cmutex
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Final answer:

To calculate the number of liters of gas formed when 450 g of NH₄NO₃ decomposes, we need to use the ideal gas law. Based on the balanced equation, we find that 5.625 moles of N₂ gas would be formed. Using the ideal gas law, we determine the volume of gas at 450 degrees C and 1.00 atm pressure to be 928.85 L.

Step-by-step explanation:

To calculate the number of liters of gas formed when 450 g of NH₄NO₃ decomposes, we need to use the ideal gas law. First, we need to find the number of moles of NH₄NO₃ by dividing the mass by the molar mass. The molar mass of NH₄NO₃ is 80 g/mol. So, the number of moles of NH₄NO₃ is 450 g / 80 g/mol = 5.625 mol.

Next, we can use the coefficients in the balanced equation to determine the number of moles of gas formed. From the equation, we see that 2 moles of NH₄NO₃ produce 2 moles of N₂, 4 moles of H₂O, and 1 mole of O₂. So, for every 2 moles of NH₄NO₃, we get 2 moles of N₂.

Since 5.625 moles of NH₄NO₃ decompose, we get 5.625 moles of N₂ gas. Finally, we can use the ideal gas law to find the volume of gas at 450 degrees C and 1.00 atm pressure. The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin. We can rearrange the equation to solve for V: V = (nRT) / P. Plugging in the values, we get V = (5.625 mol * 0.0821 L·atm/(mol·K) * (450 + 273 K)) / 1.00 atm = 928.85 L.

User Vipin Mohan
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