Question

Ammonium nitrate (NH₄NO₃) decomposes into nitrogen gas, oxygen gas, and water vapor (see reaction below). How many grams of water would get produced if 83.4 g of ammonium nitrate decomposes completely. (Assume 100% completion)

2NH₄NO₃(s) — + 2N₂(g) + O₂(g) + 4H₂O(g)

Answer 1

To find the mass of water produced when 83.4 g of ammonium nitrate decomposes completely, we use the stoichiometry of the balanced equation. For every 2 moles of ammonium nitrate, 4 moles of water are produced. Therefore, when 83.4 g of ammonium nitrate decomposes completely, 37.5 g of water is produced.

Step-by-step explanation:

To find the mass of water produced when 83.4 g of ammonium nitrate decomposes completely, we can use the stoichiometry of the balanced equation. According to the equation 2NH₄NO₃(s) → 2N₂(g) + O₂(g) + 4H₂O(g), we see that for every 2 moles of ammonium nitrate, 4 moles of water are produced. First, we calculate the moles of ammonium nitrate: 83.4 g / (80.05 g/mol) = 1.042 mol.

Then, using the stoichiometric ratio, we find the moles of water produced: 1.042 mol NH₄NO₃ * (4 mol H₂O / 2 mol NH₄NO₃) = 2.084 mol H₂O. Finally, we convert the moles of water to grams using the molar mass of water: 2.084 mol H₂O * (18.02 g/mol) = 37.5 g water.