Question

The first step in the manufacture of nitric acid by the Ostwald process is the reaction of ammonia gas with oxygen, producing nitrogen oxide and steam. The reaction mixture contains \(7.60 \mathrm{~g}\) of ammonia and \(10.00 \mathrm{~g}\) of oxygen. After the reaction is complete, \(6.22 \mathrm{~g}\) of nitrogen oxide are obtained. (a) Write a balanced equation for the reaction. (b) How many grams of nitrogen oxide can be theoretically obtained? (c) How many grams of excess reactant are theoretically unused? (d) What is the percent yield of the reaction?

Short answer

Answer: The balanced equation for the reaction between ammonia and oxygen is $$4\,\text{NH}_3\,(g) + 5\,\text{O}_2\,(g) \rightarrow 4\,\text{NO}\,(g) + 6\,\text{H}_2\text{O}\,(g)$$. The percent yield of the reaction is 82.91%.

Step-by-step solution

Write the balanced equation for the reaction

We begin by writing the balanced equation for the reaction between ammonia (NH3) and oxygen (O2), which produces nitrogen oxide (NO) and water (H2O). $$4\,\text{NH}_3\,(g) + 5\,\text{O}_2\,(g) \rightarrow 4\,\text{NO}\,(g) + 6\,\text{H}_2\text{O}\,(g)$$ The balanced equation for the reaction is given by the equation above.

Calculate the theoretical grams of nitrogen oxide

First, we'll find the moles of ammonia and oxygen using their given masses and molar masses. For ammonia, the molar mass is 17.031 g/mol, and for oxygen, the molar mass is 31.998 g/mol. Moles of ammonia = \( \frac{7.60 \, \mathrm{g}}{17.031\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.4463\, \mathrm{mol} \) Moles of oxygen = \( \frac{10.00 \, \mathrm{g}}{31.998\,\frac{\mathrm{g}}{\mathrm{mol}}} = 0.3125\, \mathrm{mol} \) Now, we'll find the limiting reactant by comparing the mole ratios. Mole ratio of ammonia to oxygen in the balanced equation is \( \frac{4}{5} = 0.8 \) Mole ratio of ammonia to oxygen in the given mixture is \( \frac{0.4463}{0.3125} = 1.428 \) Since the mole ratio in the mixture is greater than the mole ratio in the balanced equation, ammonia is in excess and oxygen is the limiting reactant. Using the stoichiometry, we calculate the theoretical moles of nitrogen oxide that can be obtained: Moles of nitrogen oxide = \( \frac{4}{5} \times 0.3125\, \mathrm{mol} = 0.2500\, \mathrm{mol} \) Finally, we convert the moles of nitrogen oxide to grams, using its molar mass (30.006 g/mol): Theoretical grams of nitrogen oxide = \( 0.2500\, \mathrm{mol} \times 30.006\, \frac{\mathrm{g}}{\mathrm{mol}} = 7.5015\, \mathrm{g} \)

Calculate the unused grams of excess reactant

Since ammonia is the excess reactant, we need to find the amount of ammonia that has reacted using the limiting reactant (oxygen). Moles of ammonia that reacted = \( \frac{4}{5} \times 0.3125\, \mathrm{mol} = 0.2500\, \mathrm{mol} \) Then, we calculate the grams of ammonia that reacted: Grams of ammonia that reacted = \( 0.2500\, \mathrm{mol} \times 17.031\, \frac{\mathrm{g}}{\mathrm{mol}} = 4.2578\, \mathrm{g} \) Now, subtract the grams of ammonia that reacted from the initial amount to find the unused grams of ammonia: Unused grams of ammonia = \( 7.60\, \mathrm{g} - 4.2578\, \mathrm{g} = 3.3422\, \mathrm{g} \)

Calculate the percent yield of the reaction

Finally, we calculate the percent yield of nitrogen oxide using the actual yield (6.22 g) and the theoretical yield (7.5015 g): Percent yield = \( \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 \) Percent yield = \( \frac{6.22 \, \mathrm{g}}{7.5015\, \mathrm{g}} \times 100 = 82.91 \% \) To summarize: a) The balanced equation for the reaction is $$4\,\text{NH}_3\,(g) + 5\,\text{O}_2\,(g) \rightarrow 4\,\text{NO}\,(g) + 6\,\text{H}_2\text{O}\,(g)$$ b) The theoretical grams of nitrogen oxide that can be obtained is 7.5015 g. c) The unused grams of excess reactant (ammonia) is 3.3422 g. d) The percent yield of the reaction is 82.91%.