Question

Ammonium nitrate (NH4NO3) is a high explosive because it is unstable and decomposes into several substances as follows: _NH4NO3--> _N2 + _O2 + _H2O. Balance the equation and then find the number of grams of each product if 42.54g of ammonium nitrate reacted.

Answer 1

The balanced reaction is : 2 NH₄NO₃ 2 N₂ + O₂ + 4 H₂O

Given : mass NH₄NO₃ reacted = 42.54 g

moles NH₄NO₃ reacted = (mass NH₄NO₃) / (molar mass NH₄NO₃)

moles NH₄NO₃ reacted = (42.54 g) / (80.04 g/mol)

moles NH₄NO₃ reacted = 0.5314 mol

(a.) moles N₂ formed = (moles NH₄NO₃ reacted) * (2 moles N₂ / 2 moles NH₄NO₃)

moles N₂ formed = (0.5314 mol) * (2 / 2)

moles N₂ formed = (0.5314 mol) * (1)

moles N₂ formed = 0.5314 mol

mass N₂ formed = (moles N₂ formed) * (molar mass N₂)

mass N₂ formed = (0.5314 mol) * (28.0 g/mol)

mass N₂ formed = 14.89 g

(b.) moles O₂ formed = (moles NH₄NO₃ reacted) * (1 moles O₂ / 2 moles NH₄NO₃)

moles O₂ formed = (0.5314 mol) * (1 / 2)

moles O₂ formed = (0.5314 mol) * (0.5)

moles O₂ formed = 0.2657 mol

mass O₂ formed = (moles O₂ formed) * (molar mass O₂)

mass O₂ formed = (0.2657 mol) * (32.0 g/mol)

mass O₂ formed = 8.50 g

(c.) moles H₂O formed = (moles NH₄NO₃ reacted) * (4 moles H₂O / 2 moles NH₄NO₃)

moles H₂O formed = (0.5314 mol) * (4 / 2)

moles H₂O formed = (0.5314 mol) * (2)

moles H₂O formed = 1.063 mol

mass H₂O formed = (moles H₂O formed) * (molar mass H₂O)

mass H₂O formed = (1.063 mol) * (18.0 g/mol)

mass H₂O formed = 19.15 g