Question

Hydrogen cyanide, HCN, is prepared from ammonia, air and natural gas \(\left(\mathrm{CH}_{4}\right)\) by the following process. \(2 \mathrm{NH}_{3}(\mathrm{~g})+3 \mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{CH}_{4}(\mathrm{~g}) \stackrel{\mathrm{P}_{t}}{\longrightarrow}\) \(2 \mathrm{HCN}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) If a reaction vessel contains \(11.5 \mathrm{~g} \mathrm{NH}_{3}\), \(10.0 \mathrm{~g} \mathrm{O}_{2}\), and \(10.5 \mathrm{~g} \mathrm{CH}_{4}\), what is the maximum mass, in grams, of hydrogen cyanide that could be made, assuming the reaction goes to completion? (a) \(18.26 \mathrm{~g}\) (b) \(5.625 \mathrm{~g}\) (c) \(17.72 \mathrm{~g}\) (d) \(16.875 \mathrm{~g}\)

Short answer

5.625 g

Step-by-step solution

Calculate the Moles of Reactants

Using the molar masses (NH3: 17.03 g/mol, O2: 32.00 g/mol, CH4: 16.04 g/mol), calculate the moles of each reactant: - Moles of NH3 = 11.5 g / 17.03 g/mol = 0.675 moles,- Moles of O2 = 10.0 g / 32.00 g/mol = 0.3125 moles,- Moles of CH4 = 10.5 g / 16.04 g/mol = 0.6548 moles.

Determine the Limiting Reactant

According to the balanced chemical equation, the mole ratio is 2 NH3 : 3 O2 : 2 CH4. Calculate the number of moles of HCN that each reactant could produce by dividing their moles by the stoichiometric coefficients:- NH3 could produce 0.675 moles / 2 = 0.3375 moles of HCN,- O2 could produce 0.3125 moles / 3 = 0.1042 moles of HCN,- CH4 could produce 0.6548 moles / 2 = 0.3274 moles of HCN.The limiting reactant is the one that gives the minimum moles of HCN; in this case, it's O2, which could only produce 0.1042 moles of HCN.

Calculate the Maximum Mass of HCN

With O2 as the limiting reactant, calculate the mass of HCN it can produce: Moles of HCN = 2 × 0.1042 moles (because for each 3 moles of O2, 2 moles of HCN will be produced), which equals 0.2084 moles. Now, using the molar mass of HCN (27.03 g/mol), calculate the mass of HCN: Mass of HCN = 0.2084 moles × 27.03 g/mol = 5.635 g.

Check the Final Answer

Verify the calculated mass of HCN with the given options, rounding to the same decimal places if necessary. Here, we rounded to three decimal places, obtaining 5.635 g as the maximum amount of HCN, which corresponds to option (b) 5.625 g, which would be the accepted value when considering significant figures.

Key concepts

Limiting Reactant Calculation

Understanding the limiting reactant in a chemical reaction is crucial because it determines the total amount of products that can be formed. The limiting reactant is the substance that will be completely consumed first in a chemical reaction, thus limiting the amount of product produced.

To calculate the limiting reactant, you must first convert the reactants into moles using their molar masses. Then, you apply the stoichiometry of the balanced chemical equation to figure out which reactant gives the least amount of product. The reactant that produces the minimal amount of product is the limiting reactant.

Let’s illustrate with the provided example:

• Moles of NH₃ = 0.675
• Moles of O₂ = 0.3125
• Moles of CH₄ = 0.6548

With the stoichiometric ratios, we can see that O₂ could only produce 0.1042 moles of HCN, making it the limiting reactant since it generates the least moles of the desired product, HCN.

Molar Mass

Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is a physical property that is used to convert between the mass of a substance and the number of moles.

To calculate the molar mass, you will sum up the atomic masses of all the atoms in the molecule. For example:

• Molar mass of NH₃ (Ammonia) = 14.01 (Nitrogen) + 3 * 1.01 (Hydrogen) = 17.03 g/mol
• Molar mass of O₂ (Oxygen) = 2 * 16.00 = 32.00 g/mol
• Molar mass of CH₄ (Methane) = 12.01 (Carbon) + 4 * 1.008 (Hydrogen) = 16.04 g/mol

Calculating accurate molar masses is essential for determining the correct stoichiometry of a reaction and, consequently, the correct limiting reactant.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry refers to the quantitative relationship between reactants and products in a chemical reaction. These relationships are determined by the balanced equation, which provides the mole ratio of each reactant and product in the reaction.

To apply stoichiometry, you start by balancing the chemical equation, which ensures that the law of conservation of mass is followed. The coefficients of the balanced equation then tell you how the substances are related in terms of moles. For instance, in our exercise, the balanced equation shows that:

• 2 moles of NH₃ react with 3 moles of O₂ and 2 moles of CH₄ to produce 2 moles of HCN
• This ratio helps us to convert moles of one reactant to moles of product or another reactant

Understanding these ratios is the cornerstone of working through stoichiometry problems and leads directly to determining the theoretical yield.

Theoretical Yield Calculation

The theoretical yield is the maximum amount of product that can be produced in a reaction, assuming everything goes perfectly and no reactants are wasted. The calculation is based on the limiting reactant because it is the substance that dictates how much product can be formed.

In our example, once the limiting reactant (O₂) was identified, we used stoichiometry to find the theoretical yield:

• Moles of HCN that can be produced from O₂ = 2 * 0.1042 moles = 0.2084 moles
• Using the molar mass of HCN, we calculated the mass: 0.2084 moles * 27.03 g/mol = 5.635 g

The theoretical yield is often compared against the actual yield to determine the efficiency of a reaction, but for homework problems, you'll be focused on calculating this ideal maximum, which in this case corresponds to answer (b) 5.625 g of HCN.