when ammonium nitrate decomposes, nitrogen, hydrogen, and oxygen gas are generated per the reaction 2NH4NO3(s)=2N2(g)+4H2+3O2(g). when 23 g NH4NO3 decomposes what total volume of gas at 1 atm and 298 do you get?
Convert 23 g NH4NO3 to mols. mols = grams/molar mass.
For every 2 mols of NH4NO3 that decomposes you obtain 2 mols N2 + 4 mols H2 + 3 mols oxygen. So you obtain 9 mols gas per 2 mols NH4NO3 or 9/2 mols gas/mol NH4NO3. After you determine total mols gas, then use PV = nRT to convert mol at STP to mols at 1 atm and 298. Post your work if you get stuck.
v=1.29(0.0821)(298)=31.56?
Looks ok to me.
To find the total volume of gas generated when 23 g of NH4NO3 decomposes, we need to calculate the moles of each gas produced, and then use the ideal gas law to find the volume.
1. Calculate the moles of NH4NO3:
Given the molar mass of NH4NO3 is 80.05 g/mol,
moles of NH4NO3 = mass / molar mass
moles of NH4NO3 = 23 g / 80.05 g/mol
moles of NH4NO3 = 0.2874 mol (approx.)
2. Use the stoichiometry of the reaction to determine the moles of N2, H2, and O2 generated:
From the balanced equation: 2NH4NO3(s) = 2N2(g) + 4H2(g) + 3O2(g)
- Ratio of NH4NO3 to N2 is 2:2 = 1:1
- Ratio of NH4NO3 to H2 is 2:4 = 1:2
- Ratio of NH4NO3 to O2 is 2:3
So, moles of N2 generated = 0.2874 mol
moles of H2 generated = 2 x 0.2874 mol = 0.5748 mol
moles of O2 generated = 3 x 0.2874 mol = 0.8622 mol
3. Now, we can use the ideal gas law equation to find the volume:
Ideal Gas Law: PV = nRT
Where P = pressure (1 atm)
V = volume (to be determined)
n = moles of gas
R = ideal gas constant (0.0821 L.atm/(mol.K))
T = temperature (298 K)
For N2:
V_N2 = (n_N2 * R * T) / P
V_N2 = (0.2874 mol * 0.0821 L.atm/(mol.K) * 298 K) / 1 atm
V_N2 = 6.823 L (approx.)
For H2:
V_H2 = (n_H2 * R * T) / P
V_H2 = (0.5748 mol * 0.0821 L.atm/(mol.K) * 298 K) / 1 atm
V_H2 = 13.656 L (approx.)
For O2:
V_O2 = (n_O2 * R * T) / P
V_O2 = (0.8622 mol * 0.0821 L.atm/(mol.K) * 298 K) / 1 atm
V_O2 = 20.526 L (approx.)
4. Total volume of gas:
Total Volume = V_N2 + V_H2 + V_O2
Total Volume ≈ 6.823 L + 13.656 L + 20.526 L
Total Volume ≈ 41.005 L
Therefore, when 23 g of NH4NO3 decomposes, the total volume of gas generated at 1 atm and 298 K is approximately 41.005 L.