Problem 63 - Chemistry: Principles and Reactions

Question

One way to remove nitrogen oxide (NO) from smoke stack 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 \(0.839\) mol of ammonia. (b) the mass of NO required to react with \(3.402\) 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.

Step-by-Step Solution

Answer

Answer: The mass of ammonia required to react with 115 g of NO is 43.5 g.

Step 1: Identify the mole ratio of ammonia to water

According to the balanced equation, the mole ratio between ammonia (NH3) and water (H2O) is 4:6 or simplified to 2:3.

Step 2: Calculate the moles of water

To determine the moles of water produced from 0.839 moles of ammonia, use the mole ratio from Step 1: $$ \text{moles of H}_2\text{O} = \frac{3}{2} \times 0.839\text{ moles of NH}_3 = 1.259\text{ moles of H}_2\text{O} $$

Step 3: Calculate the mass of water

Now, we need to convert the moles of water to mass. The molar mass of water (H2O) is approximately 18.015 g/mol. Multiply the moles of water by the molar mass of water to find the mass: $$ \text{Mass of H}_2\text{O} = 1.259 \text{ moles of H}_2\text{O} \times 18.015 \frac{\text{g}}{\text{mol}} = 22.7 \text{ g} $$ #a) The mass of water produced from 0.839 mol of ammonia is 22.7 g. #b) Mass of NO required to react with 3.402 mol of ammonia#

Step 1: Identify the mole ratio of ammonia to NO

According to the balanced equation, the mole ratio between ammonia (NH3) and nitrogen oxide (NO) is 4:6 or simplified to 2:3.

Step 2: Calculate the moles of NO required

To determine the moles of NO required to react with 3.402 moles of ammonia, use the mole ratio from Step 1: $$ \text{moles of NO} = \frac{3}{2} \times 3.402\text{ moles of NH}_3 = 5.103\text{ moles of NO} $$

Step 3: Calculate the mass of NO required

Now, we need to convert the moles of NO to mass. The molar mass of NO is approximately 30.01 g/mol. Multiply the moles of NO by the molar mass of NO to find the mass: $$ \text{Mass of NO} = 5.103 \text{ moles of NO} \times 30.01 \frac{\text{g}}{\text{mol}} = 153.1 \text{ g} $$ #b) The mass of NO required to react with 3.402 mol of ammonia is 153.1 g. #c) Mass of ammonia required to produce 12.0 g of nitrogen gas#

Step 1: Convert the mass of nitrogen gas to moles

To determine the moles of nitrogen gas (N2), we need to convert the given mass (12.0 g) to moles using the molar mass of N2 (approximately 28.02 g/mol): $$ \text{moles of N}_2 = \frac{12.0 \text{ g}}{28.02 \frac{\text{g}}{\text{mol}}} = 0.428\text{ moles of N}_2 $$

Step 2: Identify the mole ratio of ammonia to nitrogen gas

According to the balanced equation, the mole ratio between ammonia (NH3) and nitrogen gas (N2) is 4:5.

Step 3: Calculate the moles of ammonia required

To determine the moles of ammonia required to produce 0.428 moles of nitrogen gas, use the mole ratio from Step 2: $$ \text{moles of NH}_3 = \frac{4}{5} \times 0.428\text{ moles of N}_2 = 0.342\text{ moles of NH}_3 $$

Step 4: Calculate the mass of ammonia required

Now, we need to convert the moles of NH3 to mass. The molar mass of NH3 is approximately 17.03 g/mol. Multiply the moles of NH3 by the molar mass of NH3 to find the mass: $$ \text{Mass of NH}_3 = 0.342 \text{ moles of NH}_3 \times 17.03 \frac{\text{g}}{\text{mol}} = 5.82 \text{ g} $$ #c) The mass of ammonia required to produce 12.0 g of nitrogen gas is 5.82 g. #d) Mass of ammonia required to react with 115 g of NO#

Step 1: Convert the mass of NO to moles

To determine the moles of NO, we need to convert the given mass (115 g) to moles using the molar mass of NO (approximately 30.01 g/mol): $$ \text{moles of NO} = \frac{115 \text{ g}}{30.01 \frac{\text{g}}{\text{mol}}} = 3.833\text{ moles of NO} $$

Step 2: Identify the mole ratio of ammonia to NO

According to the balanced equation, the mole ratio between ammonia (NH3) and nitrogen oxide (NO) is 4:6 or simplified to 2:3.

Step 3: Calculate the moles of ammonia required

To determine the moles of ammonia required to react with 3.833 moles of NO, use the mole ratio from Step 2: $$ \text{moles of NH}_3 = \frac{2}{3} \times 3.833\text{ moles of NO} = 2.555\text{ moles of NH}_3 $$

Step 4: Calculate the mass of ammonia required

Now, we need to convert the moles of NH3 to mass. The molar mass of NH3 is approximately 17.03 g/mol. Multiply the moles of NH3 by the molar mass of NH3 to find the mass: $$ \text{Mass of NH}_3 = 2.555 \text{ moles of NH}_3 \times 17.03 \frac{\text{g}}{\text{mol}} = 43.5 \text{ g} $$ #d) The mass of ammonia required to react with 115 g of NO is 43.5 g.

Key Concepts

Mole Ratio

The mole ratio is a foundational concept in stoichiometry that tells us how many moles of one substance react or are produced in relation to another substance in a chemical reaction. Think of the mole ratio as a recipe for a chemical equation; it guides us in mixing the correct amounts of reactants to get our desired products.

In our exercise example, the balanced chemical equation shows that 4 moles of ammonia react with 6 moles of nitrogen oxide to produce 5 moles of nitrogen gas and 6 moles of water. This ratio (4:6:5:6) is crucial for computations like determining the mass of water produced from a known amount of ammonia. By using a simplified mole ratio (2:3 for ammonia to water), students can calculate the mass of other reactants or products involving these species. Correctly identifying and applying the mole ratio is key to successful stoichiometry calculations.

Molar Mass

Molar mass is the weight of one mole of a substance, expressed in grams per mole (g/mol). This number is unique to each chemical compound and is equal to the sum of the atomic weights of all atoms in the compound's formula. It serves as a bridge between the mass of a substance and the number of moles.

To find the molar mass, look up the atomic weights of each element on the periodic table and add them up according to the number of atoms in the compound's formula. In the exercise, for water (H₂O), we calculate its molar mass by summing the atomic weights of two hydrogen atoms (about 1.01 g/mol each) and one oxygen atom (about 16.00 g/mol), which gives approximately 18.015 g/mol for water. This conversion factor is then used to translate moles of a substance, like ammonia or water, into grams, allowing students to understand the amount of material involved in the chemical reaction.

Chemical Reaction Quantification

Quantifying chemical reactions is the act of determining the exact amounts of reactants and products involved in a chemical reaction. Be it the mass of gases emitted from a smokestack or the production in an industrial process, stoichiometry allows us to calculate these quantities.

Following the balanced chemical equation and using mole ratios and molar masses gives us the power to compute the exact amounts needed or produced. For example, when given the quantity of one reactant, like ammonia, one can figure out how much NO is needed for complete reaction or how much water will be produced. This concept sets the stage for predicting outcomes in real-world chemical processes, such as environmental engineering to reduce pollutants, thus optimizing the use of resources and reducing waste.