2NH4NO3 --> 2N2 + O2 + 4H2O

What mass of H2O can be formed from the decomposition of 12.0 g of NH4NO3?

To determine the mass of H2O formed from the decomposition of NH4NO3, we need to use stoichiometry.

1. Start by writing down the balanced equation:
2NH4NO3 → 2N2 + O2 + 4H2O

2. Calculate the molar mass of NH4NO3 (ammonium nitrate):
NH4NO3 = (1 × 1) + (4 × 1) + (3 × 16) = 80 g/mol

3. Convert the given mass of NH4NO3 to moles:
Moles of NH4NO3 = Mass / Molar mass = 12.0 g / 80 g/mol = 0.15 mol

4. Use the mole ratio between NH4NO3 and H2O to find the moles of H2O produced:
Moles of H2O = Moles of NH4NO3 × (4 moles H2O / 2 moles NH4NO3) = 0.15 mol × (4/2) = 0.30 mol

5. Finally, calculate the mass of H2O using the molar mass of water:
Molar mass of H2O = (2 × 1) + (16 × 1) = 18 g/mol
Mass of H2O = Moles of H2O × Molar mass of H2O = 0.30 mol × 18 g/mol = 5.4 g

Therefore, from the decomposition of 12.0 g of NH4NO3, the mass of H2O formed is 5.4 g.