Answer:
70% is the percent yield
Explanation:
First of all, we work with the stoichiometry of the decomposition:
NH₄NO₃ (s) → N₂O (g) + 2H₂O (l)
1 mol of ammonium nitrate can produce 1 mol of nitrogen oxide.
This, will happen with the 100% yield reaction.
To determine the percent yield we do:
(moles produced / theoretical yield) . 100 =
(0.7 moles / 1 mol) . 100 = 70%