Problem 74 - Chemistry: The Central Science

Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

Step-by-Step Solution

Answer

(a) The limiting reactant is O₂. (b) 2.062 g of NO and 1.858 g of H₂O form. (c) 0.33 g of NH₃ remain after the limiting reactant is completely consumed. (d) The total mass of reactants (4.25 g) equals the total mass of products and remaining reactant (4.25 g), which is consistent with the law of conservation of mass.

Step 1: Calculate the moles of reactants

First, we need to calculate the moles of both ammonia and oxygen gas using their given masses and molar masses. The molar mass of NH₃ is 14.01 g/mol (for N) + 3 * 1.01 g/mol (for H) = 17.034 g/mol, and the molar mass of O₂ is 2 * 16.00 g/mol = 32.00 g/mol. Moles of NH₃ = mass of NH₃ / Molar mass of NH₃ = 1.50 g / 17.034 g/mol = 0.0881 mol Moles of O₂ = mass of O₂ / Molar mass of O₂ = 2.75 g / 32.00 g/mol = 0.0859 mol

Step 2: Determine the limiting reactant

Looking at the balanced equation for the reaction, we notice that 4 moles of NH₃ react with 5 moles of O₂. To determine the limiting reactant, we will compare the mole ratios with the stoichiometry of the reaction: Mole ratio of NH₃ to O₂ = (0.0881 mol NH₃) / (0.0859 mol O₂) = 1.0256 Since this value is greater than the stoichiometry ratio (4/5 = 0.8), the limiting reactant is O₂ because there is not enough O₂ to react with all the NH₃.

Step 3: Calculate the grams of products formed

Now we will calculate the grams of NO and H₂O formed using the limiting reactant (O₂) and the stoichiometry of the reaction. From the balanced equation, 5 moles of O₂ produce 4 moles of NO and 6 moles of H₂O. So, Moles of NO formed = (4/5) * moles of O₂ = (4/5) * 0.0859 mol = 0.0687 mol Moles of H₂O formed = (6/5) * moles of O₂ = (6/5) * 0.0859 mol = 0.1031 mol Now convert moles of the products to grams using their molar masses. The molar mass of NO is 14.01 g/mol (for N) + 16.00 g/mol (for O) = 30.01 g/mol, and the molar mass of H₂O is 2 * 1.01 g/mol (for H) + 16.00 g/mol (for O) = 18.02 g/mol. Grams of NO formed = moles of NO * Molar mass of NO = 0.0687 mol * 30.01 g/mol = 2.062 g Grams of H₂O formed = moles of H₂O * Molar mass of H₂O = 0.1031 mol * 18.02 g/mol = 1.858 g

Step 4: Calculate the remaining mass of the excess reactant

We have already identified NH₃ as the excess reactant. Now, we will calculate how much NH₃ has reacted and will determine the remaining mass of NH₃ after the reaction. Moles of NH₃ reacted = (4/5) * moles of O₂ = (4/5) * 0.0859 mol = 0.0687 mol Grams of NH₃ reacted = moles of NH₃ reacted * Molar mass of NH₃ = 0.0687 mol * 17.034 g/mol = 1.17 g Remaining mass of NH₃ = initial mass of NH₃ – mass of NH₃ reacted = 1.50 g – 1.17 g = 0.33 g

Step 5: Verify the consistency of calculations with the law of conservation of mass

According to the law of conservation of mass, the total mass of the reactants should be equal to the total mass of the products and any remaining reactants. So we have: Total mass of reactants = mass of NH₃ + mass of O₂ = 1.50 g + 2.75 g = 4.25 g Total mass of products and remaining reactant = grams of NO formed + grams of H₂O formed + remaining mass of NH₃ = 2.062 g + 1.858 g + 0.33 g = 4.25 g Since the total mass of reactants equals the total mass of products and remaining reactant, our calculations are consistent with the law of conservation of mass.

Key Concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculation of reactants and products in chemical reactions. It is essential for determining how much of each substance is needed or produced in a reaction. In the provided exercise, we use stoichiometry to find out which reactant will limit the production of products (limiting reactant) and to calculate the amount of nitric oxide (NO) and water (H₂O) generated.

The stoichiometry of a reaction is dictated by the coefficients in the balanced chemical equation. For example, in the conversion of ammonia (\(\text{NH}_3\)) to nitric oxide and water, the balanced equation is:\[4\ \text{NH}_{3}(g) + 5\ \text{O}_{2}(g) \rightarrow 4\ \text{NO}(g) + 6\ \text{H}_{2}\text{O}(g)\]These coefficients (4:5:4:6) tell us that 4 moles of ammonia react with 5 moles of oxygen to produce 4 moles of nitric oxide and 6 moles of water. This ratio is crucial for determining how much of each reagent is involved in the reaction.

To utilize stoichiometry:

• Calculate the moles of each reactant using their given weights and molar masses.
• Use the coefficients from the balanced equation to compare the actual mole ratio of the reactants.
• Identify the limiting reactant by determining which reactant's mole ratio is less than required.
• Calculate the moles and subsequently the grams of products formed using the stoichiometric coefficients and the moles of the limiting reactant.

Mastering stoichiometry is key to successfully predicting the quantities of substances in chemical reactions.

Law of Conservation of Mass

The Law of Conservation of Mass is a core principle in chemistry stating that mass is neither created nor destroyed in a chemical reaction. This means that the mass of reactants at the beginning of a reaction will equal the mass of products and any leftover reactants when the reaction is complete.

In the given exercise, it is verified by comparing the total mass of the reactants with the total mass of the products and remaining excessive reactants. Initially, the calculation involves adding the mass of ammonia (NH₃) and oxygen (O₂) to ensure they sum up to the mass of nitrogen monoxide (NO), water (H₂O), and any excess ammonia that hasn't reacted.

Here's how you can check this:

• Add the initial masses of NH₃ and O₂: 1.50 g + 2.75 g = 4.25 g.
• Calculate the mass of NO and H₂O produced and add any remaining mass of excess NH₃: 2.062 g (NO) + 1.858 g (H₂O) + 0.33 g (remaining NH₃) = 4.25 g.

When these totals are equal, as shown in the exercise, the calculations are consistent with the law. This principle is reliable because it applies universally to all chemical reactions, assuring that no mass is lost or mysteriously gained.

Understanding and applying the Law of Conservation of Mass helps ensure that calculations involving chemical reactions are both accurate and meaningful.

Chemical Equations

Chemical equations are symbolic representations of chemical reactions, displaying the reactants and products along with their quantities in mole units. A balanced chemical equation is essential because it follows the Law of Conservation of Mass, showing equal total atoms of each element on both sides of the equation.

In the case of ammonia reacting with oxygen in the exercise:\[4 \ \text{NH}_{3}(g) + 5 \ \text{O}_{2}(g) \rightarrow 4 \ \text{NO}(g) + 6 \ \text{H}_{2}\text{O}(g)\]This equation is balanced because it reflects that all atoms present in the reactants are accounted for in the products:

• Ammonia (NH₃) is represented by 4 molecules.
• Each oxygen molecule is counted 5 times.
• Nitric oxide (NO) and water (H₂O) have 4 and 6 molecules respectively.

Balancing chemical equations requires skill and practice as it involves ensuring that the same number of each type of atom is found on both sides of an equation. This ensures no molecules are lost, symbolizing the reaction’s progress accurately.

Moreover, writing balanced chemical equations helps chemists provide insightful predictions regarding the amounts of reactants needed and quantities of products obtained in a reaction, vital for both theoretical studies and practical applications.