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) The limiting reactant is \(\mathrm{O}_{2}\). (b) 1.88 g of \(\mathrm{NO}\) and 1.69 g of \(\mathrm{H}_{2}\mathrm{O}\) form. (c) 0.928 g of \(\mathrm{NH}_{3}\) remain. (d) The mass of reactants (4.50 g) is approximately equal to the mass of products and excess reactant (4.50 g), which is consistent with the law of conservation of mass.

Step 1: Calculate the moles of NH3 and O2

We need to determine the number of moles of NH3 and O2 from their given mass. Use the formula: moles = mass (in grams) / molar mass (in g/mol). Molar mass of \(\mathrm{NH}_{3}\) is 17.03 g/mol, and for \(\mathrm{O}_{2}\), it is 32.00 g/mol. Number of moles of \(\mathrm{NH}_{3} =\frac{2.00\,\mathrm{g}}{17.03\,\mathrm{g/mol}} =0.117\,\mathrm{mol}\) Number of moles of \(\mathrm{O}_{2} =\frac{2.50\,\mathrm{g}}{32.00\,\mathrm{g/mol}} =0.0781\,\mathrm{mol}\)

Step 2: Identify the limiting reactant

To find the limiting reactant, we need to examine the stoichiometric ratios in the balanced equation. We will compare the ratios of the initial moles to the coefficients in the balanced equation. Ratios: \(\frac{\mathrm{mol\,NH}_{3}}{4} = \frac{0.117}{4} = 0.0292\) \(\frac{\mathrm{mol\,O}_{2}}{5} = \frac{0.0781}{5} = 0.0156\) Since the ratio of oxygen is smaller than that of ammonia, \(\mathrm{O}_{2}\) is the limiting reactant.

Step 3: Calculate the mass of NO and H2O

Now that we know the limiting reactant, we can calculate the mass of the products \(\mathrm{NO}\) and \(\mathrm{H}_{2}\mathrm{O}\). Use stoichiometric relationships (coefficients in the balanced equation) to find the moles of products formed by the limiting reactant. Moles of \(\mathrm{NO}\) formed: \(\mathrm{mol\, O}_{2} \times(\frac{4\,\text{mol}\,\mathrm{NO}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{4}{5})=0.0625\,\text{mol}\,\mathrm{NO}\) Moles of \(\mathrm{H}_{2}\mathrm{O}\) formed: \(\mathrm{mol\, O}_{2} \times(\frac{6\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{6}{5})=0.0936\,\text{mol}\,\mathrm{H}_{2}\mathrm{O}\) We can now convert the moles of products to grams by multiplying with the molar mass of the respective product: Mass of \(\mathrm{NO} = 0.0625\,\mathrm{mol} \times 30.01\,\mathrm{g/mol} = 1.88\,\mathrm{g}\) Mass of \(\mathrm{H}_{2}\mathrm{O} = 0.0936\,\mathrm{mol} \times 18.02\,\mathrm{g/mol} = 1.69\,\mathrm{g}\)

Step 4: Calculate the mass of the excess reactant remaining

Now that we found our limiting reactant and the products formed, we can compute the mass of the excess reactant, \(\mathrm{NH}_{3}\), that will remain after the limiting reactant is completely consumed. Moles of \(\mathrm{NH}_{3}\) consumed: \(\mathrm{mol\, O}_{2} \times(\frac{4\,\text{mol}\,\mathrm{NH}_{3}}{5\,\text{mol}\,\mathrm{O}_{2}})= 0.0781 \times(\frac{4}{5})=0.0625\,\text{mol}\,\mathrm{NH}_{3}\) Moles of \(\mathrm{NH}_{3}\) remaining: \(0.117 \, \mathrm{mol} - 0.0625 \, \mathrm{mol} = 0.0545 \, \mathrm{mol}\) Mass of \(\mathrm{NH}_{3}\) remaining: \(0.0545\,\mathrm{mol} \times 17.03\,\mathrm{g/mol} = 0.928\, \mathrm{g}\)

Step 5: Verify the law of conservation of mass

To check if our calculations are consistent with the law of conservation of mass, we will ensure that the mass of the reactants equals the mass of the products and the excess reactant remaining. Mass of reactants: \(2.00\,\mathrm{g\,of\, NH}_{3}+2.50\,\mathrm{g\,of\, O}_{2}=4.50\,\mathrm{g}\) Mass of products and excess reactant: \(1.88\,\mathrm{g\,of\, NO} + 1.69\,\mathrm{g\,of\, H}_{2}\mathrm{O} + 0.928\,\mathrm{g\,of\, NH}_{3} \approx 4.50\,\mathrm{g}\) The mass of reactants and products are approximately equal, which is consistent with the law of conservation of mass.

Key Concepts

Stoichiometric Calculations

Stoichiometric calculations are at the heart of predicting the outcomes of chemical reactions. They allow us to quantitatively understand the relation between the amounts of reactants and products in a chemical equation. To perform such calculations, we start by identifying the balanced chemical equation and then use the coefficients of the reactants and products, which are known as stoichiometric coefficients, to find out the molar ratios between them.

These coefficients indicate the relative amounts of each substance involved in the reaction. For instance, a balanced equation with coefficients 2A + 3B → 4C means that 2 moles of A react with 3 moles of B to produce 4 moles of C. In practical terms, through stoichiometric calculations, if a student knows the amount of A, they can find out how much of B is needed or how much of C will be produced, thereby informing how to measure reactants for a given product yield, or vice versa.

Molecular Mass

Molecular mass, also known as molecular weight, is a vital concept to grasp for stoichiometric calculations. It represents the sum of the atomic masses of all the atoms in a molecule, measured in atomic mass units (amu) or grams per mole (g/mol). For instance, to calculate the molecular mass of water (H₂O), we add the atomic mass of hydrogen (roughly 1.01 amu) twice and oxygen (approx. 16.00 amu) once, resulting in approximately 18.02 amu or g/mol.

Knowing the molecular mass of a compound allows us to convert between the mass of a substance in grams and the number of moles, enabling precise stoichiometric calculations. This conversion is pivotal in identifying the limiting reactant and computing the amounts of products formed in a chemical reaction.

Law of Conservation of Mass

The law of conservation of mass is a fundamental principle in chemistry that states that mass is neither created nor destroyed in a chemical reaction. It indicates the importance of a balanced chemical equation where the total mass of the reactants must equal the total mass of the products.

When we perform stoichiometric calculations, verifying the results against this law assures us that the reaction is balanced and the calculations are accurate. For example, in a reaction where 2.00 grams of A react with 3.00 grams of B, and our calculations predict 4.50 grams of product C and 0.50 grams of unreacted B, we can be confident in our calculations if the sum of the mass of the products and unreacted reactants equals the initial mass of the reactants, 5.00 grams in this case.

Chemical Reaction Balancing

Balancing a chemical reaction is the process of ensuring that the number of atoms of each element is the same on both sides of the equation. This is imperative because it reflects the law of conservation of mass, indicating that no atoms are lost or gained during a chemical reaction.

The correct balance of a chemical equation is necessary for accurate stoichiometric calculations and to identify the limiting reactant. Each coefficient in the balanced equation provides the mole ratio of reactants to products, which is used to calculate how much of each reactant is needed and the amounts of products that will form. Balancing chemical equations is an essential skill for students to master to predict the outcomes of real-world chemical processes.