Question

ammonia (nh₃) chemically reacts with oxygen gas (o₂) to produce nitric oxide (no) and water (h₂o). what mass of ammonia is consumed by the reaction of 1.6 g of oxygen gas? round your answer to 2 significant digits.

Explanation:

Step1: Write the balanced chemical equation

The reaction between ammonia (\(NH_3\)) and oxygen (\(O_2\)) to produce nitric oxide (\(NO\)) and water (\(H_2O\)) is:

$$4NH_3 + 5O_2 ightarrow 4NO + 6H_2O$$

Step2: Calculate moles of \(O_2\)

Molar mass of \(O_2 = 2 \times 16.00 = 32.00 \, g/mol\).
Moles of \(O_2 = \frac{\text{mass of } O_2}{\text{molar mass of } O_2} = \frac{1.6 \, g}{32.00 \, g/mol} = 0.05 \, mol\).

Step3: Relate moles of \(O_2\) to \(NH_3\) (mole ratio)

From the balanced equation, the mole ratio of \(NH_3\) to \(O_2\) is \(4:5\).
Let moles of \(NH_3 = n\). Then:
$$\frac{n}{0.05 \, mol} = \frac{4}{5}$$
$$n = 0.05 \, mol \times \frac{4}{5} = 0.04 \, mol$$

Step4: Calculate mass of \(NH_3\)

Molar mass of \(NH_3 = 14.01 + 3 \times 1.01 = 17.04 \, g/mol\).
Mass of \(NH_3 = \text{moles} \times \text{molar mass} = 0.04 \, mol \times 17.04 \, g/mol \approx 0.68 \, g\) (rounded to 2 significant digits).

Answer:

\(0.68 \, g\)