Question

One way to remove nitrogen oxide (NO) from smokestack emissions is to react it with ammonia. $$ 4 \mathrm{NH}_{3}(g)+6 \mathrm{NO}(g) \longrightarrow 5 \mathrm{~N}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) $$ Calculate (a) the mass of water produced from \(0.839 \mathrm{~mol}\) of ammonia. (b) the mass of NO required to react with \(3.402 \mathrm{~mol}\) of ammonia. (c) the mass of ammonia required to produce \(12.0 \mathrm{~g}\) of nitrogen gas. (d) the mass of ammonia required to react with \(115 \mathrm{~g}\) of NO

Short answer

(a) Based on the calculation, moles of water = (0.839/4) x 6 = 1.2585 Now, mass of water = moles of water x molar mass of water mass of water = 1.2585 x 18.02 = 22.67 g So, 22.67 g of water is produced. (b) Using stoichiometry, moles of NO = (0.839/4) x 6 = 1.2585 Now, mass of NO = moles of NO x molar mass of NO mass of NO = 1.2585 x 30.01 = 37.77 g Therefore, 37.77 g of NO is required. (c) Firstly, converting the mass of N2 into moles, moles of N2 = 12.0 / 28.02 = 0.4283 Now, moles of ammonia = (0.4283/5) x 4 = 0.3426 Then, mass of ammonia = moles of ammonia x molar mass of ammonia mass of ammonia = 0.3426 x 17.03 = 5.834 g So, 5.834 g of ammonia is required. (d) Converting the mass of NO into moles, moles of NO = 115 / 30.01 = 3.8324 Now, moles of ammonia = (3.8324/6) x 4 = 2.5550 Then, mass of ammonia = moles of ammonia x molar mass of ammonia mass of ammonia = 2.5550 x 17.03 = 43.51 g Hence, 43.51 g of ammonia is required.

Step-by-step solution

(a) Calculate the mass of water produced

To do this, first, we need to find the moles of water resulting from the given moles of ammonia. Then, we'll convert the moles of water into mass using the molar mass of water. According to the balanced equation, 4 moles of ammonia react to produce 6 moles of water. So, let's find the moles of water produced from 0.839 moles of ammonia by using the stoichiometry. $$ \mathrm{moles\:of\:water} = \frac{\mathrm{moles\:of\:ammonia}}{4} \times 6 = \frac{0.839}{4} \times 6 $$ Now, use the molar mass of water (18.02 g/mol) to convert the moles of water into mass. $$ \mathrm{mass\:of\:water} = \mathrm{moles\:of\:water} \times \mathrm{molar\:mass\:of\:water} $$ Calculate and report the mass of the water produced.

(b) Calculate the mass of NO required

Similarly, we'll use the stoichiometry of the reaction to find the moles of NO required to react with the given moles of ammonia, and then convert the moles of NO into mass using the molar mass of NO. From the balanced equation, 4 moles of ammonia react with 6 moles of NO. $$ \mathrm{moles\:of\:NO} = \frac{\mathrm{moles\:of\:ammonia}}{4} \times 6 = \frac{3.402}{4} \times 6 $$ Now, use the molar mass of NO (30.01 g/mol) to convert the moles of NO into mass. $$ \mathrm{mass\:of\:NO} = \mathrm{moles\:of\:NO} \times \mathrm{molar\:mass\:of\:NO} $$ Calculate and report the mass of NO required.

(c) Calculate the mass of ammonia required

First, we need to find the moles of ammonia required to produce the given mass of nitrogen gas. Then, we'll convert the moles of ammonia into mass using the molar mass of ammonia. Use the molar mass of N2 (28.02 g/mol) to convert the mass of N2 into moles. $$ \mathrm{moles\:of\:N_{2}} = \frac{\mathrm{mass\:of\:N_{2}}}{\mathrm{molar\:mass\:of\:N_{2}}} =\frac{12.0}{28.02} $$ From the balanced equation, 5 moles of N2 are produced from 4 moles of ammonia. $$ \mathrm{moles\:of\:ammonia} = \frac{\mathrm{moles\:of\:N_{2}}}{5} \times 4 = \frac{12.0}{28.02} \times \frac{4}{5} $$ Now, use the molar mass of ammonia (17.03 g/mol) to convert the moles of ammonia into mass. $$ \mathrm{mass\:of\:ammonia} = \mathrm{moles\:of\:ammonia} \times \mathrm{molar\:mass\:of\:ammonia} $$ Calculate and report the mass of ammonia required.

(d) Calculate the mass of ammonia required

Similarly, we'll use the molar mass of NO to convert the given mass of NO into moles and find the moles of ammonia required to react with the given moles of NO. Then, convert the moles of ammonia into mass using the molar mass of ammonia. First, convert the mass of NO into moles using its molar mass, 30.01 g/mol. $$ \mathrm{moles\:of\:NO} = \frac{\mathrm{mass\:of\:NO}}{\mathrm{molar\:mass\:of\:NO}} =\frac{115}{30.01} $$ From the balanced equation, 6 moles of NO react with 4 moles of ammonia. $$ \mathrm{moles\:of\:ammonia} = \frac{\mathrm{moles\:of\:NO}}{6} \times 4 = \frac{115}{30.01} \times \frac{4}{6} $$ Now, use the molar mass of ammonia (17.03 g/mol) to convert the moles of ammonia into mass. $$ \mathrm{mass\:of\:ammonia} = \mathrm{moles\:of\:ammonia} \times \mathrm{molar\:mass\:of\:ammonia} $$ Calculate and report the mass of ammonia required.

Key concepts

Chemical Reaction Stoichiometry

Understanding chemical reaction stoichiometry is crucial when it comes to predicting the outcomes of reactions, including how much of each substance will be used up and what quantities of products will be formed. It's based on the law of conservation of mass, which states that in a chemical reaction, the total mass of the reactants is equal to the total mass of the products. In other words, atoms are neither created nor destroyed, they simply rearrange to form new substances.

When you look at a balanced chemical equation, such as the reaction between nitrogen oxide (NO) and ammonia (NH₃), the coefficients before each chemical formula represent the number of moles of that substance involved in the reaction. For instance, in the balanced reaction $$4 \text{NH}_{3}(g) + 6 \text{NO}(g) \rightarrow 5 \text{N}_{2}(g) + 6 \text{H}_{2} \text{O}(l)$$we see that 4 moles of NH₃ react with 6 moles of NO to produce 5 moles of N₂ and 6 moles of H₂O. This ratio allows us to predict the amounts of reactants and products using stoichiometry.

Let's say we have a certain amount of ammonia and we want to know how much water will be produced. Using the coefficients from the balanced equation, we can set up a ratio that will help us convert moles of ammonia into moles of water. Then, with the moles of water, we can use the molar mass of water to find the corresponding mass. This is an example of how stoichiometry serves as the quantitative relationship between reactants and products in a chemical reaction.

Molar Mass Calculation

Molar mass is a physical property defined as the mass of a given substance (chemical element or chemical compound) divided by the amount of substance. The unit for molar mass is grams per mole (g/mol), and it reflects how much one mole of a substance weighs. Calculating the molar mass of a molecule requires the addition of the atomic masses of every atom in the molecule as found on the periodic table. For example, water (H₂O) has a molar mass of approximately 18.02 g/mol because it is composed of two hydrogen atoms (approximately 1.01 g/mol each) and one oxygen atom (approximately 16.00 g/mol).

In stoichiometry problems, understanding the molar mass is key to converting between mass and moles. The formula for this conversion is:$$\text{number of moles} = \frac{\text{mass}}{\text{molar mass}}$$This formula is essential when you're given a mass of a substance and need to determine the number of moles to use in stoichiometric calculations. Conversely, if you have the number of moles and need to find the mass, you multiply the number of moles by the molar mass. Knowing the molar mass enables us to make these conversions and thus solve various parts of a stoichiometry problem.

Stoichiometric Calculations

Stoichiometric calculations are the calculations involved in balancing a chemical equation. They enable chemists to determine the relative quantities of reactants and products required or produced in a chemical reaction. To perform these calculations, one needs to use the balanced equation as a reference for the mole ratio between reactants and products.

Let's apply this to an example: if we need to calculate the mass of ammonia NH₃ required to react with a certain mass of NO, we should first convert the given mass of NO to moles using the molar mass of NO. Next, we use the stoichiometry of the balanced equation to find the moles of NH₃ that react with the moles of NO we just found. Finally, we convert the moles of NH₃ back to mass by using its molar mass.

This process involves three main steps:

• Converting mass to moles (using molar mass)
• Using the mole ratio from the balanced equation to find the number of moles of the other substance
• Converting moles back to mass, if required (using molar mass again)

By mastering these stoichiometric calculations, students can confidently tackle problems where they're asked to predict the amounts of products formed or reactants required, as demonstrated in the exercise's step-by-step solutions.