Problem 80 - 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, \(2.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.50 \mathrm{~g}\) of \(\mathrm{O}_{2}\). (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and \(\mathrm{H}_{2} \mathrm{O}\) 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) O2 is the limiting reactant. b) 1.876 g of NO and 1.689 g of H2O are formed. c) 0.935 g of NH3 remains. d) The law of conservation of mass is satisfied, as both the mass of reactants and products is 4.50 g.

Step 1: Calculate moles of NH3

To calculate moles of NH3, use the formula: moles = (mass of NH3) / (molar mass of NH3) The molar mass of NH3 is 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.03 g/mol. moles of NH3 = (2.00 g) / (17.03 g/mol) = 0.1174 mol

Step 2: Calculate moles of O2

To calculate moles of O2, use the formula: moles = (mass of O2) / (molar mass of O2) The molar mass of O2 is 2 * 16.00 g/mol = 32.00 g/mol. moles of O2 = (2.50 g) / (32.00 g/mol) = 0.07813 mol #Step 2: Determine the limiting reactant#

Step 3: Compare mole ratios

Use the balanced equation to determine the mole ratio between NH3 and O2: 4 NH3 : 5 O2 Divide the moles of each substance by their respective stoichiometric coefficient: NH3: (0.1174 mol) / 4 = 0.02935 O2: (0.07813 mol) / 5 = 0.01563 Since 0.01563 (O2) is lower than 0.02935 (NH3), O2 is the limiting reactant. #Step 3: Calculate the amounts of NO and H2O formed#

Step 4: Calculate moles of NO formed

Using the stoichiometric coefficients in the balanced equation, calculate the moles of NO formed: 4 moles of NH3 form 4 moles of NO. Since O2 is the limiting reactant, we determine the moles of NO using the O2 mole ratio: (0.07813 mol O2) * (4 mol NO / 5 mol O2) = 0.06250 mol NO

Step 5: Calculate grams of NO formed

Now, convert moles of NO to grams using its molar mass, 30.01 g/mol: (0.06250 mol) * (30.01 g/mol) = 1.876 g NO

Step 6: Calculate moles of H2O formed

Following the balanced equation, calculate the moles of H2O formed: 4 moles of NH3 form 6 moles of H2O. Using the limiting reactant, O2's mole ratio: (0.07813 mol O2) * (6 mol H2O / 5 mol O2) = 0.09375 mol H2O

Step 7: Calculate grams of H2O formed

Convert moles of H2O to grams using its molar mass,18.02 g/mol: (0.09375 mol) * (18.02 g/mol) = 1.689 g H2O #Step 4: Determine the remaining amount of excess reactant#

Step 8: Calculate moles of NH3 remaining

We already know the amount of NH3 reacted with O2: (0.07813 mol O2) * (4 mol NH3 / 5 mol O2) = 0.06250 mol NH3 Now subtract the moles of NH3 reacted from the initial moles of NH3: 0.1174 mol - 0.06250 mol = 0.0549 mol NH3 remaining

Step 9: Calculate grams of NH3 remaining

Convert moles of NH3 remaining to grams using its molar mass, 17.03 g/mol: (0.0549 mol) * (17.03 g/mol) = 0.935 g NH3 remaining #Step 5: Verify consistency with the law of conservation of mass#

Step 10: Calculate the mass of reactants

Calculate the combined mass of NH3 and O2 initially given: mass of reactants = 2.00 g NH3 + 2.50 g O2 = 4.50 g

Step 11: Calculate total mass of products

Calculate the combined mass of NO, H2O, and remaining NH3: mass of products = 1.876 g NO + 1.689 g H2O + 0.935 g NH3 remaining = 4.50 g Since the mass of reactants is equal to the mass of products (4.50 g), our calculations are consistent with the law of conservation of mass.

Key Concepts

Limiting Reactant

Understanding the concept of the limiting reactant is essential for students to solve stoichiometry problems efficiently. It's like being at a hot dog stand: if you have 10 hot dog buns and only 8 hot dogs, you can only make 8 complete hot dogs. The hot dogs are the limiting reactant.

In a chemical reaction, the limiting reactant is the substance that will be completely used up first, determining the maximum amount of product that can be formed. The reaction stops when there is no more limiting reactant, and any other reactants present in excess remain unreacted.

Identifying the Limiting Reactant

To identify the limiting reactant, follow these steps:

• Convert the mass of each reactant to moles using the molar mass of each compound.
• Use the balanced chemical equation to find the mole ratio between reactants.
• Divide the moles of each reactant by its coefficient in the balanced equation.
• The reactant with the smaller result is the limiting reactant.

Applying this method to the exercise about ammonia and oxygen shows that oxygen (O₂) is the limiting reactant.

Mole-to-Mass Conversion

The mole-to-mass conversion is a fundamental skill in stoichiometry, enabling chemists to translate between the microscopic world of atoms and molecules and the macroscopic world we can measure. The molar mass serves as the bridge between moles, which count particles, and mass in grams.

Performing Mole-to-Mass Conversions

Let’s break down how to perform these conversions with an example from the exercise:

• Determine the molar mass of the substance by adding up the atomic masses of all atoms in the molecule.
• Calculate the number of moles by dividing the given mass by the molar mass of the substance.
• To find the mass from the number of moles, multiply the number of moles by the molar mass.

For instance, to find the mass of NO produced, we first calculate the moles of NO from the limiting reactant, O₂, and then convert those moles to mass.

Law of Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry, stating that in a closed system, mass is neither created nor destroyed in a chemical reaction; it is conserved. This law underpins all stoichiometry calculations.

To ensure calculations adhere to this law, the sum of the mass of the reactants should always equal the sum of the mass of the products. If they don't, something went wrong with the calculations or measurements.

Applying Conservation of Mass

In our exercise, the initial masses of ammonia (NH₃) and oxygen (O₂) add up to 4.50 grams. After reacting, the total mass of nitrogen monoxide (NO), water (H₂O), and the remaining ammonia should also equal 4.50 grams to follow the conservation law. This step is crucial to validate the work done in previous steps and ensure the reaction equation correctly represents the conservation of mass.