Question

1. ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4nh₃(g) + 5o₂(g) → 4no(g) + 6h₂o(g) what is the theoretical yield of h₂o, in grams, when 40.0 g nh₃ and 50.0 g o₂ are mixed and allowed to react? (molar masses: nh₃ = 17.03 g/mol ; o₂ = 32.0 g/mol; h₂o = 18.0 g/mol) a. 63.4 g b. 52.8 g c. 33.8 g d. 21.3 g e. 68.2 g

Explanation:

Step1: Calculate moles of reactants

Moles of \(NH_3=\frac{40.0\ g}{17.03\ g/mol}\approx2.35\ mol\)
Moles of \(O_2=\frac{50.0\ g}{32.0\ g/mol}=1.5625\ mol\)

Step2: Determine limiting reactant

From the balanced equation \(4NH_3 + 5O_2
ightarrow4NO + 6H_2O\), the mole ratio of \(NH_3\) to \(O_2\) is \(\frac{4}{5}\).
If all \(NH_3\) reacts, \(O_2\) needed \(n(O_2)=\frac{5}{4}\times2.35\ mol = 2.9375\ mol\) (more than available).
If all \(O_2\) reacts, \(NH_3\) needed \(n(NH_3)=\frac{4}{5}\times1.5625\ mol = 1.25\ mol\) (less than available). So \(O_2\) is the limiting reactant.

Step3: Calculate moles of \(H_2O\)

Using the mole ratio of \(O_2\) to \(H_2O\) (\(\frac{5}{6}\)), moles of \(H_2O=\frac{6}{5}\times1.5625\ mol = 1.875\ mol\)

Step4: Calculate mass of \(H_2O\)

Mass of \(H_2O=n\times M = 1.875\ mol\times18.0\ g/mol = 33.75\ g\approx33.8\ g\)

Answer:

C. 33.8 g