Question

One mol of ammonia reacts with 1.00 mol of oxygen to form nitrogen oxide and water according to the reaction $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ State which statements are true about the reaction and make the false statements true. (a) All the oxygen is consumed. (b) \(4.00 \mathrm{~mol} \mathrm{NO}\) are produced. (c) \(1.50 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) are produced. (d) The description of the experiment does not provide enough information to determine percent yield. (e) Three moles of water are produced for every two moles of NO obtained.

Short answer

Question: Analyze the following statements about a reaction where 1.00 mol of ammonia (NH3) and 1.00 mol of oxygen (O2) are mixed and react to form nitrogen oxide (NO) and water (H2O). Label each statement as true or false. If a statement is false, correct it to make it true. a) All the oxygen is consumed. b) 4.00 mol NO are produced. c) 1.50 mol H2O are produced. d) The description of the experiment does not provide enough information to determine percent yield. e) Three moles of water are produced for every two moles of NO obtained. Answer: a) False. Corrected statement: Not all the oxygen is consumed. b) False. Corrected statement: 1.00 mol NO is produced. c) True d) True e) True

Step-by-step solution

Statement (a): All the oxygen is consumed.

According to the given reaction, we have: $$4 \mathrm{NH}_3(g) + 5 \mathrm{O}_2(g) \longrightarrow 4 \mathrm{NO}(g) + 6 \mathrm{H}_2 \mathrm{O}(l)$$ The stoichiometric ratio between \(\mathrm{NH}_3\) and \(\mathrm{O}_2\) is 4:5. Since there is 1.00 mol of both ammonia and oxygen, we can see that there will be some oxygen left over after the reaction. Hence, this statement is false. To make this statement true, we can say: "Not all the oxygen is consumed."

Statement (b): 4.00 mol NO are produced.

The stoichiometric ratio between \(\mathrm{NH}_3\) and \(\mathrm{NO}\) is 1:1. Since 1.00 mol of ammonia is used, 1.00 mol of nitrogen oxide will be produced, therefore this statement is false. To make this statement true, we can say: "1.00 mol \(\mathrm{NO}\) is produced."

Statement (c): 1.50 mol H2O are produced.

The stoichiometric ratio between \(\mathrm{NH}_3\) and \(\mathrm{H}_2\mathrm{O}\) is 2:3. Since 1.00 mol of ammonia is used, according to the stoichiometric ratio, we can calculate the amount of water produced: $$1.00\,\mathrm{mol}\,\mathrm{NH}_3 \times \frac{3\,\mathrm{mol}\,\mathrm{H}_2 \mathrm{O}}{2\,\mathrm{mol}\,\mathrm{NH}_3} = 1.50\,\mathrm{mol}\,\mathrm{H}_2 \mathrm{O}$$ Hence, statement (c) is true.

Statement (d): The description of the experiment does not provide enough information to determine percent yield.

Since we are not given information about the actual amount of products formed or their masses, we cannot determine the percent yield of the reaction. Therefore, statement (d) is true.

Statement (e): Three moles of water are produced for every two moles of NO obtained.

The stoichiometric ratio between \(\mathrm{NO}\) and \(\mathrm{H}_2\mathrm{O}\) is 4:6, which can be simplified to 2:3. This means that for every 2 moles of nitrogen oxide, 3 moles of water are produced. Hence, statement (e) is true.