Question

In the reaction, \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) When \(1 \mathrm{~mol}\) of ammonia and \(1 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) are made to react to completion then: (a) \(1.0 \mathrm{~mol}\) of \(\mathrm{H}_{2} \mathrm{O}\) is produced (b) \(1.0 \mathrm{~mol}\) of \(\mathrm{NO}\) will be produced (c) All the ammonia will be consumed (d) All the oxygen will be consumed

Short answer

(c) is true since not all of the ammonia is consumed; (d) is true as all the oxygen is consumed.

Step-by-step solution

Write Down the Chemical Reaction

The given chemical reaction is: \[4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\] This means 4 moles of \(\mathrm{NH}_3\) react with 5 moles of \(\mathrm{O}_2\) to produce 4 moles of \(\mathrm{NO}\) and 6 moles of \(\mathrm{H}_2\mathrm{O}\).

Determine Limiting Reactant

Start with 1 mole of \(\mathrm{NH}_3\) and 1 mole of \(\mathrm{O}_2\). Since 4 moles of \(\mathrm{NH}_3\) should react with 5 moles of \(\mathrm{O}_2\), the ratio is \(\frac{4}{5}\) for \(\mathrm{NH}_3\) to \(\mathrm{O}_2\). Calculate moles of oxygen needed for 1 mole of ammonia:\[\text{Moles of } \mathrm{O}_2 = \frac{5}{4} \times 1 = 1.25 \text{ moles}\]Since we only have 1 mole of \(\mathrm{O}_2\), \(\mathrm{O}_2\) is the limiting reactant.

Calculate Amount of Products Formed

Since \(\mathrm{O}_2\) is the limiting reactant, calculate how much \(\mathrm{NO}\) and \(\mathrm{H}_2\mathrm{O}\) will be produced from 1 mole of \(\mathrm{O}_2\):From the balanced equation, - 5 moles of \(\mathrm{O}_2\) produce 4 moles of \(\mathrm{NO}\).- 5 moles of \(\mathrm{O}_2\) produce 6 moles of \(\mathrm{H}_2 \mathrm{O}\).For \(\mathrm{NO}\):\[\text{Moles of } \mathrm{NO} = \frac{4}{5} \times 1 = 0.8 \text{ moles}\]For \(\mathrm{H}_2\mathrm{O}:\)\[\text{Moles of } \mathrm{H}_2 \mathrm{O} = \frac{6}{5} \times 1 = 1.2 \text{ moles}\]

Analyze the Consumption of Reactants

Since \(\mathrm{O}_2\) is the limiting reactant, it will be completely consumed. From the equation, \(5 \text{ moles of } \mathrm{O}_2\) require \(4 \text{ moles of } \mathrm{NH}_3\). The moles of \(\mathrm{NH}_3\) that will react with 1 mole of \(\mathrm{O}_2\) are:\[\text{Moles of } \mathrm{NH}_3 = \frac{4}{5} \times 1 = 0.8 \text{ moles}\]This means only 0.8 moles of \(\mathrm{NH}_3\) will be consumed, and we initially had 1 mole, so \(\mathrm{NH}_3\) will not be completely consumed.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative relationships in chemical reactions. At its core, stoichiometry is about calculating the amounts of reactants and products involved in a chemical reaction. It allows us to predict how much of each substance is required to react or will be produced. By using the balanced chemical equation, we can set up ratios that help us determine these quantities.

In the example of the reaction \(4 \text{NH}_3(\text{g})+5 \text{O}_2(\text{g}) \rightarrow 4 \text{NO}(\text{g})+6 \text{H}_2\text{O}(\text{l})\), we start with knowing that 4 moles of ammonia \(\text{NH}_3\) need to react with 5 moles of oxygen \(\text{O}_2\). This means for every 4 molecules of ammonia, we need 5 molecules of oxygen. This ratio helps us calculate which reactant runs out first and hence becomes the limiting reactant.

• This concept is key to understanding how much of a product can be formed in any chemical reaction.
• Stoichiometry is not just limited to molecules and moles; it involves understanding how masses of substances are involved in reactions too, through molar mass conversions.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. Reactions involve breaking and forming chemical bonds, which changes the arrangement of atoms. Understanding chemical reactions is essential for solving problems related to chemical processes and predicting the outcomes of chemical mixes.

In our specific exercise, ammonia \(\text{NH}_3\) reacts with oxygen \(\text{O}_2\) to produce nitrogen monoxide \(\text{NO}\) and water \(\text{H}_2\text{O}\). Inside this reaction, you can see how molecules interact:

• The bonds in collective reactant molecules break.
• New bonds are formed, yielding product molecules.

The balanced equation simplifies this interaction by showing the proportion of reactants to products. It is essential to remember that all chemical reactions, in theory, are reversible, but conditions such as temperature and pressure can influence their progression and direction.

Balanced Equation

A balanced chemical equation is crucial for correctly reflecting a chemical reaction. It obeys the Law of Conservation of Mass, which states that matter cannot be created or destroyed in a chemical reaction. Thus, the number of each type of atom is the same on both sides of the equation.

In our reaction, \(4 \text{NH}_3(\text{g})+5 \text{O}_2(\text{g}) \rightarrow 4 \text{NO}(\text{g})+6 \text{H}_2\text{O}(\text{l})\), balancing ensures that:

• Every element appears equally on both sides, maintaining atomic balance.
• We can use these balanced coefficients (like 4 moles \(\text{NH}_3\) and 5 moles \(\text{O}_2\)) in stoichiometric calculations.

Understanding balanced equations allows us to identify limiting reactants and calculate how much of the products will be formed. This is crucial for accurately predicting the results of a chemical reaction and applying this knowledge to practical laboratory settings.