Problem 63 - Chemistry

Question

(a) What volumes (in liters) of ammonia and oxygen must react to form \(12.8 \mathrm{~L}\) of nitric oxide according to the equation at the same temperature and pressure? $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ (b) What volumes (in liters) of propane and water vapor must react to form \(8.96 \mathrm{~L}\) of hydrogen according to the equation at the same temperature and pressure? $$ \mathrm{C}_{3} \mathrm{H}_{8}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) \longrightarrow 3 \mathrm{CO}(g)+7 \mathrm{H}_{2}(g) $$

Step-by-Step Solution

Answer

For part (a), 12.8 L of \(NH_{3}\) and 16 L of \(O_{2}\) are required. For part (b), 1.254 L of \(C_{3}H_{8}\) and 3.853 L of \(H_{2}O\) are required.

Step 1: Establish Mole Ratios for the First Reaction

From the balanced chemical equation \(4NH_{3}(g) + 5O_{2}(g) \rightarrow 4NO(g) + 6H_{2}O(g)\), we can establish that 4 volumes of \(NH_{3}\), and 5 volumes of \(O_{2}\), react to form 4 volumes of \(NO\). This means ratio of the volume of \(NH_{3}\) to the volume of \(NO\) is 1 (i.e., 4/4), and the ratio of the volume of \(O_{2}\) to the volume of \(NO\) is 1.25 (i.e., 5/4).

Step 2: Calculate Volumes for the First Reaction

We know \(12.8 L\) of \(NO\) are formed. So using the volume ratios established in step 1, the volume of \(NH_{3}\) required would be \(12.8 L * 1 = 12.8 L\), and the volume of \(O_{2}\) required would be \(12.8 L * 1.25 = 16 L\).

Step 3: Establish Mole Ratios for the Second Reaction

From the balanced chemical equation \(C_{3}H_{8}(g) + 3H_{2}O(g) \rightarrow 3CO(g) + 7H_{2}(g)\), we can establish that 1 volume of \(C_{3}H_{8}\), and 3 volumes of \(H_{2}O\), react to form 7 volumes of \(H_{2}\). This means the ratio of the volume of \(C_{3}H_{8}\) to the volume of \(H_{2}\) is 0.14 (i.e., 1/7), and the ratio of the volume of \(H_{2}O\) to the volume of \(H_{2}\) is 0.43 (i.e., 3/7).

Step 4: Calculate Volumes for the Second Reaction

We know \(8.96 L\) of \(H_{2}\) are formed. So using the volume ratios established in step 3, the volume of \(C_{3}H_{8}\) required would be \(8.96 L * 0.14 = 1.254 L\), and the volume of \(H_{2}O\) required would be \(8.96 L * 0.43 = 3.853 L\).

Key Concepts

Understanding Volume Ratios in Reactions

In chemistry, particularly in stoichiometry, volume ratios help us understand how much of each reactant is needed in a gaseous chemical reaction, or how much product we can form.Volume ratios come from the coefficients of the balanced equation, which illustrate the number of moles needed for the reaction. In gases, due to Avogadro's law, these mole ratios translate directly into volume ratios when the gases are at the same temperature and pressure.For instance, in the reaction between ammonia (\(\mathrm{NH}_3\)) and oxygen (\(\mathrm{O}_2\)) to produce nitric oxide (\(\mathrm{NO}\)), the equation says:{linebreak}\[4\,\mathrm{NH}_3(g) + 5\,\mathrm{O}_2(g) \rightarrow 4\,\mathrm{NO}(g) + 6\,\mathrm{H}_2\mathrm{O}(g)\]{linebreak}We deduce that:

• 4 volumes of \(\mathrm{NH}_3\) react with 5 volumes of \(\mathrm{O}_2\)
• This forms 4 volumes of \(\mathrm{NO}\)

This gives a straightforward volume ratio of 1:1 for \(\mathrm{NH}_3\) to \(\mathrm{NO}\), and a ratio of 1.25:1 for \(\mathrm{O}_2\) to \(\mathrm{NO}\) because 5 divided by 4 equals 1.25.

Basics of Chemical Reactions

Chemical reactions involve the transformation of substances through breaking and forming chemical bonds. In a balanced chemical equation, the number of each type of atom is the same on both sides of the equation.The reaction of propane (\(\mathrm{C}_3\mathrm{H}_8\)) with steam (\(\mathrm{H}_2\mathrm{O}\)) demonstrates stoichiometric principles. The balanced equation is:{linebreak}\[\mathrm{C}_3\mathrm{H}_8(g) + 3\,\mathrm{H}_2\mathrm{O}(g) \rightarrow 3\,\mathrm{CO}(g) + 7\,\mathrm{H}_2(g)\]{linebreak}Here’s how we interpret this:

• 1 molecule (or unit) of \(\mathrm{C}_3\mathrm{H}_8\) reacts with 3 molecules of \(\mathrm{H}_2\mathrm{O}\)
• This forms 3 molecules of \(\mathrm{CO}\) and 7 molecules of \(\mathrm{H}_2\)

Understanding these numbers shows us that the ratios in our equation dictate how much of each reactant is needed, and how much product will be formed.Balanced equations allow us to predict and calculate these amounts accurately, showcasing the law of conservation of mass.

Explaining Gas Laws in Chemistry

Gas laws are fundamental in stoichiometric calculations involving gaseous substances. They describe the behavior of gases in relation to temperature, volume, and pressure. One of the key laws is Avogadro's Law, which states that "equal volumes of gases at the same temperature and pressure contain the same number of molecules." This principle allows mole ratios in chemical equations to be treated as volume ratios when gases are involved. In stoichiometry, when identities of gases and their conditions remain constant, we use these relationships:

• Given the volumes of gases at a uniform temperature and pressure, the coefficients in a balanced chemical equation directly translate into volume ratios.
• Boyle’s Law and Charles’s Law further guide how changing conditions would affect gas volumes.

For activities at constant conditions, the simplifying assumption often used is that the volumes of reactants and products align exactly as the coefficients in a balanced equation suggest. Through applying these laws, complex calculations become more manageable, and we gain better insight into how changes in conditions affect reaction progress.