Question

How many moles of oxygen (O₂) are needed to react with 56.8 grams of ammonia?

4 NH3 + 5 O2 → 4 NO + 6 H2O 

Answer 1

M(NH₃) = 17.03 g/mol

n(NH₃) $= \frac{m}{M} = \frac{56.8}{17.03} = 3.335 \;mol$

According to the reaction:

n(O₂) = $\frac{5}{4}$ n(NH₃) = $\frac{5}{4} \times 3.335 = 4.17 \;mol$

Answer: 4.17 mol