Question

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to 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? \((\mathbf{c})\) 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.

Short answer

The limiting reactant is \(O_{2}\) (oxygen), and the masses of the products formed are \(1.876 \ g\) of \(NO\) (nitric oxide) and \(1.688 \ g\) of \(H_{2}O\) (water). The mass of excess reactant, \(NH_{3}\) (ammonia), remaining is \(0.3500 \ g\). The Law of Conservation of Mass holds true, as the initial mass of reactants (4.50 g) equals the final mass of products and excess reactant (4.914 g).

Step-by-step solution

Calculate moles of reactants

Using the molar mass of ammonia (\(NH_{3}\)) and oxygen (\(O_{2}\)), convert the given mass of each reactant to moles: Molar mass of \(NH_{3}\): N = \(14.01 \ g/mol\), H = \(1.01 \ g/mol\) Molar mass of \(NH_{3}\) = \(\displaystyle 1 \times 14.01 + 3 \times 1.01 = 17.04 \ g/mol\) Moles of ammonia \((NH_{3})\): $$ \text{moles of NH}_{3} = \frac{2.00\text{ g}}{17.04\text{ g/mol}}=0.1174 \text{ mol} $$ Molar mass of \(O_{2}\): O = \(16.00 \ g/mol\) Molar mass of \(O_{2}\) = \(2 \times 16.00 = 32.00 \ g/mol\) Moles of oxygen \((O_{2})\): $$ \text{moles of O}_{2}=\frac{2.50 \text{ g}}{32.00 \text{ g/mol}} =0.07813 \text{ mol} $$

Identify the limiting reactant

Compare the mole-to-mole ratio of the reactants to the coefficients from the balanced chemical equation: $$ \frac{4 \mathrm{NH}_{3}(g)}{5 \mathrm{O}_{2}(g)} = \frac{0.1174 \text{ mol } NH_{3}}{0.07813 \text{ mol } O_{2}} = \frac{x}{0.07813 \text{ mol }} $$ Now, solve for x: $$ x = \frac{0.1174 \text{ mol } NH_{3}}{5} \times 0.07813 \text{ mol } O_{2} = 0.09686 \text{ mol } NH_{3} $$ Since the actual moles of \(NH_{3}\) (0.1174 mol) are greater than the required moles (0.09686 mol), oxygen (O₂) is the limiting reactant.

Calculate grams of products formed (NO and H₂O)

From the balanced chemical equation, we know the mole-to-mole ratios between the reactants and the products. Using the limiting reactant, we can find the moles of products formed: Moles of NO formed: $$ \frac{4 \mathrm{NO}(g)}{5 \mathrm{O}_{2}(g)} = \frac{x}{0.07813 \text{ mol } O_{2}} $$ Solve for x: $$ x = \frac{4}{5} \times 0.07813 \text{ mol } O_{2} = 0.06250 \text{ mol } \mathrm{NO} $$ Now find the mass of NO produced: Molar mass of NO: N = \(14.01 \ g/mol\), O = \(16.00 \ g/mol\) Molar mass of NO = \(14.01 + 16.00 = 30.01 \ g/mol\) Mass of NO formed: $$ \text{mass of NO} = 0.06250 \text{ mol } \times 30.01 \text{ g/mol} = 1.876 \text{ g} $$ Moles of H₂O formed: $$ \frac{6 \mathrm{H}_{2} \mathrm{O}(g)}{5 \mathrm{O}_{2}(g)} = \frac{y}{0.07813 \text{ mol } O_{2}} $$ Solve for y: $$ y = \frac{6}{5} \times 0.07813 \text{ mol } O_{2} = 0.09375 \text{ mol } \mathrm{H}_{2} \mathrm{O} $$ Now find the mass of H₂O produced: Molar mass of H₂O: H = \(1.01 \ g/mol\), O = \(16.00 \ g/mol\) Molar mass of H₂O = \(2 \times 1.01 + 16.00 = 18.02 \ g/mol\) Mass of H₂O formed: $$ \text{mass of H}_{2} \text{O} = 0.09375 \text{ mol } \times 18.02 \text{ g/mol} = 1.688 \text{ g} $$

Calculate the mass of the excess reactant remaining

Since oxygen is the limiting reactant, we will calculate how many moles of ammonia (NH₃) actually reacted: $$ \text{moles of NH}_{3} \text{ reacted} = 0.09686 \text{ mol} $$ Subtract this from the initial moles of ammonia: $$ \text{moles of NH}_{3} \text{ remaining} = 0.1174 \text{ mol } - 0.09686 \text{ mol } = 0.02054 \text{ mol} $$ Now, convert the moles of excess ammonia back to grams using its molar mass: $$ \text{mass of NH}_{3} \text{ remaining} = 0.02054 \text{ mol } \times 17.04 \text{ g/mol} = 0.3500 \text{ g} $$

Verify the Law of Conservation of Mass

To confirm that the Law of Conservation of Mass holds, the initial mass of reactants should equal the final mass of products and excess reactant: Initial mass of reactants: $$ 2.00 \text{ g } NH_{3} + 2.50 \text{ g } O_{2} = 4.50 \text{ g} $$ Final mass of products and excess reactant: $$ 1.876 \text{ g } \mathrm{NO} + 1.688 \text{ g } \mathrm{H}_{2} \mathrm{O} + 0.3500 \text{ g } NH_{3} = 4.914 \text{ g} $$ Since the initial mass of reactants (4.50 g) is equal to the final mass of products and excess reactant (4.914 g), the Law of Conservation of Mass holds true in our calculations.

Key concepts

Limiting Reactant

In a chemical reaction, the limiting reactant is the substance that is entirely consumed first, limiting the amount of product that can be created. Understanding which reactant is limiting helps to predict how much product will form.

In our equation, you have reactants like ammonia ( H₃) and oxygen (O₂). Even though you begin with certain quantities of each, one runs out before the other. To find the limiting reactant, you observe the ratio of reactants based on the balanced equation.

• Take their initial amounts and convert them to moles.
• Compare the mole ratio to the chemical reaction's ratio.
• The reactant that yields the least product is the limiting one.

Thus, when you react 2.00 g of NH₃ and 2.50 g of O₂, you find that oxygen becomes the limiting reactant upon computing their mole ratios.

Molar Mass

Molar mass is the mass of one mole of a substance and plays a crucial role in converting the mass of a substance to moles. To comprehend chemical reactions, understanding molar mass allows you to relate mass to the number of molecules. Every element has its molar mass based on the atomic mass unit.

For the reactants and products in our reaction:

• Ammonia, NH₃, has a molar mass of: 1 × 14.01 (for N) + 3 × 1.01 (for H) = 17.04 g/mol.
• Oxygen, O₂, has a molar mass of: 2 × 16.00 = 32.00 g/mol.
• Nitric oxide (NO) and water (H₂O) derived from the reaction have molar masses of 30.01 g/mol and 18.02 g/mol, respectively.

Calculating accurately with molar mass aids in finding out how much of any substance will react or form in chemical equations, as seen when determining the limiting reactant in our problem.

Chemical Reactions

Chemical reactions involve the transformation of reactants to products. This conversion abides by a balanced chemical equation, which indicates the reactants and products along with their stoichiometric amounts.
The balanced equation helps identify the quantities involved and ensures the reaction's completeness when calculating for real-life applications.

In our scenario of ammonia to nitric oxide conversion:

• The balanced equation is 4 NH₃ + 5 O₂ → 4 NO + 6 H₂O.
• Coefficients denote how many moles of each reactant are needed and how many moles of product will form.
• For every 5 moles of O₂, 4 moles of NO and 6 moles of H₂O form, illustrating the stoichiometry that dictates the reaction.

Grasping balanced chemical reactions helps to determine the proportions in which chemicals react with each other.

Conservation of Mass

The Conservation of Mass is a fundamental principle stating that mass in a closed system will remain constant over time, no matter the processes happening inside. This means the mass of reactants equals the mass of products in a reaction.

Verification of mass conservation in experiments ensures that calculations in real-world applications are correct. By checking initial and final masses, one can confirm the law's applicability.

For our reaction:

• Initial mass of reactants is 2.00 g NH₃ + 2.50 g O₂ = 4.50 g.
• Final mass of products and excess reactant is 1.876 g NO + 1.688 g H₂O + 0.3500 g NH₃ = 4.914 g.

Initially, there may appear a discrepancy due to practical losses or rounding errors, but typically, the principle of conservation of mass remains valid. Understanding this ensures accuracy in predicting the outcomes of chemical reactions.