Question

Nitrogen oxide is a pollutant commonly found in smokestack emissions. One way to remove it is to react it with ammonia. 4NH3(g) 1 6NO(g) 9: 5N2(g) 1 6H2O(l) How many liters of ammonia are required to change 12.8 L of nitrogen oxide to nitrogen gas? Assume 100% yield and that all gases are measured at the same temperature and pressure.

Short answer

Answer: 8.53 liters of ammonia are required to react with 12.8 liters of nitrogen oxide, assuming 100% yield and all gases are measured at the same temperature and pressure.

Step-by-step solution

Write down the balanced chemical equation

We are given the balanced chemical equation as: 4NH3(g) + 6NO(g) → 5N2(g) + 6H2O(l)

Calculate the mole ratio

Next, we need to determine the mole ratio between NH3 and NO from the balanced chemical equation. We see that it is: 4 moles NH3 : 6 moles NO

Apply ratio to volume

Apply the mole ratio to the volume ratio (since all gases are measured at the same temperature and pressure) to find the volume of NH3 needed to react with 12.8 L NO. _(Volume NH3) / (Volume NO) = (4 moles NH3) / (6 moles NO)_

Solve for the volume of ammonia

Now, solve for the volume of NH3 needed to react with 12.8 L NO. _(Volume NH3) = (4 moles NH3) / (6 moles NO) × (12.8 L NO)_ _Volume NH3 = (4/6) × 12.8 L_ _Volume NH3 = 8.53 L_ (rounded to two decimal places) So, 8.53 liters of ammonia are required to change 12.8 liters of nitrogen oxide to nitrogen gas, assuming 100% yield and all gases are measured at the same temperature and pressure.

Key concepts

Balanced Chemical Equation

A balanced chemical equation is crucial in understanding the stoichiometry of a chemical reaction. It represents the conservation of mass principle, indicating that atoms are neither created nor destroyed in a chemical reaction. To balance an equation, one must ensure that the number of atoms for each element is the same on both sides of the reaction.

In the context of the nitrogen oxide and ammonia reaction, the balanced chemical equation is:
\[ 4NH_3(g) + 6NO(g) \rightarrow 5N_2(g) + 6H_2O(l) \]
This equation tells us the exact proportions in which reactants combine and products form. Without a balanced equation, making accurate predictions about the amounts of reactants needed or products formed would be impossible. It provides the foundation upon which mole ratios and gas volume calculations are made, making it a cornerstone of stoichiometry in chemical reactions.

Mole Ratio

The mole ratio is derived from the coefficients of a balanced chemical equation and serves as a conversion factor between moles of reactants and products. In the given reaction, the mole ratio between ammonia (NH3) and nitrogen oxide (NO) can be interpreted directly from the coefficients in the balanced equation as follows:
\[4 moles\text{ }NH_3 : 6 moles\text{ }NO\]
Therefore, for every 6 moles of NO reacting, 4 moles of NH3 are required. These ratios are essential for converting between amounts of different substances involved in a reaction. By understanding mole ratios, students can solve various stoichiometry problems, such as determining the amount of one reactant needed to completely react with a given quantity of another reactant.

Gas Volume Calculations

In gas volume calculations, we apply the ideal gas law principle which states that at constant temperature and pressure, equal volumes of gases contain an equal number of moles. Thus, we can use the mole ratio previously determined to calculate the volume of one gas required to react with a given volume of another.

Using the exercise as an example, to find the volume of NH3 needed to react with 12.8 L of NO, we apply the mole ratio between NH3 and NO, alongside the given volumes:
\[ \frac{\text{Volume NH}_3}{\text{Volume NO}} = \frac{4\text{ moles NH}_3}{6\text{ moles NO}} \]
By inputting the known volume of NO we can solve for the required volume of NH3:
\[ \text{Volume NH}_3 = \left( \frac{4}{6} \right) \times 12.8\text{ L NO} \]
\[ \text{Volume NH}_3 = 8.53\text{ L} \]
It is crucial to note that this calculation assumes that the gases behave ideally and are at the same temperature and pressure, which is often the case in stoichiometry problems. This technique can be widely applied to calculate volumes in gas-gas reactions.