Ammonium nitrate has been used as high explosive because its unstable & decomposes into several gaseous substances. The rapid expansion of the gaseous substances produces the explosive force.
NH4NO3-->N2+O2_H2O
Calculate the mass of each product gas if 1.25g of ammonium nitrate reacts.
To calculate the mass of each product gas, we first need to determine the balanced equation for the reaction of ammonium nitrate.
The balanced equation is:
NH4NO3 → N2 + O2 + H2O
According to the equation, one molecule of ammonium nitrate (NH4NO3) decomposes into one molecule of nitrogen gas (N2), one molecule of oxygen gas (O2), and one molecule of water (H2O).
Next, we need to calculate the molar masses of each compound:
- Molar mass of NH4NO3:
N (14.01 g/mol) x 2 + H (1.01 g/mol) x 4 + O (16.00 g/mol) x 3 = 80.04 g/mol
Now, we can calculate the number of moles of NH4NO3:
Number of moles = Mass / Molar mass
Number of moles = 1.25 g / 80.04 g/mol ≈ 0.0156 mol
Since the balanced equation shows a 1:1:1:1 molar ratio between NH4NO3, N2, O2, and H2O, we can conclude that the number of moles of each product gas will also be 0.0156 mol.
Finally, we can calculate the mass of each product gas by multiplying the number of moles by their respective molar masses:
- Mass of N2: 0.0156 mol x (28.01 g/mol) = 0.436 g
- Mass of O2: 0.0156 mol x (32.00 g/mol) = 0.499 g
- Mass of H2O: 0.0156 mol x (18.02 g/mol) = 0.281 g
Therefore, the mass of each product gas (N2, O2, H2O) when 1.25 g of ammonium nitrate reacts is approximately:
- Mass of N2: 0.436 g
- Mass of O2: 0.499 g
- Mass of H2O: 0.281 g
To calculate the mass of each product gas when 1.25g of ammonium nitrate (NH4NO3) reacts, we need to first determine the molar masses of each compound involved.
The molar mass of ammonium nitrate (NH4NO3) can be calculated by summing the molar masses of each element present in the compound. Taking into account nitrogen (N), hydrogen (H), oxygen (O), and the atomic masses from the periodic table, we get:
Molar mass of NH4NO3 = (4 x (1.01 g/mol)) + (1 x (14.01 g/mol)) + (3 x (16.00 g/mol))
= 28.04 g/mol + 14.01 g/mol + 48.00 g/mol
= 80.05 g/mol
Now, using the balanced chemical equation for the decomposition of ammonium nitrate (NH4NO3) into nitrogen (N2), oxygen (O2), and water (H2O), which is:
NH4NO3 --> N2 + O2 + H2O
We can see that for every one mole of ammonium nitrate (NH4NO3), we get one mole of nitrogen (N2), one mole of oxygen (O2), and one mole of water (H2O).
Given that we have 1.25g of ammonium nitrate, we can convert this to moles using the molar mass we calculated earlier:
Moles of NH4NO3 = mass / molar mass
= 1.25g / 80.05 g/mol
≈ 0.0156 mol
Since the moles of the reactant and products are the same (as seen from the balanced equation), we have:
Moles of N2 = 0.0156 mol
Moles of O2 = 0.0156 mol
Moles of H2O = 0.0156 mol
To find the mass of each product gas, we can simply multiply the moles of each product by their respective molar masses:
Mass of N2 = Moles of N2 x molar mass of N2
= 0.0156 mol x 28.01 g/mol (molar mass of N2)
≈ 0.436 g
Mass of O2 = Moles of O2 x molar mass of O2
= 0.0156 mol x 32.00 g/mol (molar mass of O2)
≈ 0.499 g
Mass of H2O = Moles of H2O x molar mass of H2O
= 0.0156 mol x 18.02 g/mol (molar mass of H2O)
≈ 0.281 g
Therefore, when 1.25g of ammonium nitrate reacts, the mass of nitrogen (N2) produced is approximately 0.436g, the mass of oxygen (O2) produced is approximately 0.499g, and the mass of water (H2O) produced is approximately 0.281g.
Balance the equation.
Convert 1.25 g NH4NO3 to moles.
Convert moles NH4NO3 to moles N2, moles O2, and moles H2O. (Three conversions each done separately).
Convert moles of each to grams of each.