Problem 50 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

Hydrogen cyanide, \(\mathrm{HCN}\), can be made by a two-step process. First, ammonia is reacted with \(\mathrm{O}_{2}\) to give nitric oxide, \(\mathrm{NO}\). \(4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 4 \mathrm{NO}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Then nitric oxide is reacted with methane, \(\mathrm{CH}_{4}\) \(2 \mathrm{NO}(\mathrm{g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \rightarrow 2 \mathrm{HCN}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) \(+\mathrm{H}_{2}(\mathrm{~g})\) When \(25.5 \mathrm{~g}\) of ammonia and \(32.0 \mathrm{~g}\) of methane are used, how many grams of hydrogen cyanide can be produced? (a) \(1.5\) (b) \(2.0\) (c) \(40.5\) (d) \(54.0\)

Step-by-Step Solution

Answer

2.0 grams

Step 1: Calculate moles of reactants

Use the molar mass of ammonia (NH3) and methane (CH4) to convert their masses to moles. Molar masses: NH3 = 17.03 g/mol, CH4 = 16.04 g/mol. Moles of NH3 = 25.5 g / 17.03 g/mol, Moles of CH4 = 32.0 g / 16.04 g/mol.

Step 2: Determine the limiting reactant

Use stoichiometry to find which reactant will be consumed first. The stoichiometry of the reactions shows that 4 moles of NH3 produce 4 moles of NO, which further reacts with 2 moles of CH4 to produce 2 moles of HCN. Compare the calculated moles of NH3 and CH4 with the stoichiometric ratios to find the limiting reactant.

Step 3: Calculate moles of HCN produced

Using the stoichiometry of the reaction with the limiting reactant, calculate the moles of HCN that can be produced. This will be directly proportional to the moles of the limiting reactant.

Step 4: Convert moles of HCN to grams

The molar mass of HCN is 27.03 g/mol. Use this to convert the calculated moles of HCN to grams: grams of HCN = moles of HCN * 27.03 g/mol.

Step 5: Identify the answer from the options

Check the calculated grams of HCN against the provided multiple-choice options to identify the correct answer.

Key Concepts

Understanding Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is a fundamental concept in chemistry that involves the calculation of the quantities of reactants and products in a chemical reaction. It's based on the law of conservation of mass, where the total mass of reactants equals the total mass of products. In stoichiometry, we use the balanced chemical equation as a recipe to understand how different molecules react in fixed ratios, known as stoichiometric coefficients.

To successfully tackle stoichiometry problems, you should first write out and balance the chemical equation. Once balanced, you can see the mole ratios of reactants to products, which are essential in determining how much of each substance is involved in the reaction. In the exercise provided, the reaction between ammonia and oxygen to produce nitric oxide and water is balanced, which provides the basis for further calculations. Understanding this mole ratio is like knowing the recipe for a cake – if you need twice as much of one ingredient, you'll likely need twice as much of another to keep the cake's integrity.

Navigating Limiting Reactant Calculations

When solving stoichiometry problems, one key step is identifying the limiting reactant, which is the reactant that will be entirely consumed first and thus limits the amount of product that can be formed. Think of it as having a certain number of toy cars but a limited number of wheels. If you run out of wheels, you can't make any more complete cars, no matter how many car bodies you have left.

To calculate the limiting reactant, convert the mass of each reactant to moles using the molar mass. Then, compare the ratio of moles available to the ratio required by the balanced equation. The reactant with the lesser ratio available than required is the limiting reactant. In our exercise example, the moles of ammonia and methane are compared to their stoichiometric ratios to find out which one limits the production of hydrogen cyanide. Identifying the limiting reactant is crucial since it determines the maximum amount of product that can be formed.

Mastering Molar Mass Conversion

Molar mass conversion is the bridge between the mass of a substance and the number of moles, a unit that chemists use to express amounts of a chemical substance. Molar mass is defined as the mass of one mole of a substance and is expressed in grams per mole (g/mol). It can be calculated by adding up the atomic masses of the elements in the molecule, as found on the periodic table.

For example, to compute the molar mass of ammonia (NH3), simply add the mass of one nitrogen atom and three hydrogen atoms. When you're given the mass of a substance in a problem, divide the mass by the molar mass to convert it to moles. This step is showcased in our exercise where we convert the given masses of ammonia and methane into moles for further stoichiometric calculations. Mastering the conversion between mass and moles is essential for solving stoichiometry problems because stoichiometric coefficients in a balanced equation refer to moles, not grams.