How many moles of H₂ are needed to produce 9.33 moles NH₃?

Balanced equation:

N₂ + 3 H₂ ⟶ 2 NH₃

Theoretical number of moles of H₂ needed to produce 9.33 moles NH₃ :

(9.33 moles NH₃) * (3 moles H₂ / 2 moles NH₃) = 14.0 moles H₂

---

What mass in g of copper is produced when 3.8 g of Fe is reacted with excess CuSO₄ ?

Balanced equation:

Fe + CuSO₄ ⟶ Cu + FeSO₄

Theoretical yield in grams of Cu from 3.8 g Fe :

(3.8 g Fe) * (1 mole Fe / 55.85 g Fe) * (1 mole Fe / 1 mole Cu) *

(63.55 g Cu / 1 mole Cu)

= 4.3 g Cu