Question

Hydrogen cyanide is prepared commercially by the reaction of methane, \(\mathrm{CH}_{4}(\mathrm{~g})\), ammonia, \(\mathrm{NH}_{3}(\mathrm{~g})\), and oxygen, \(\mathrm{O}_{2}(\mathrm{~g})\), at high temperature. The other product is gaseous water. a. Write a chemical equation for the reaction. b. What volume of \(\mathrm{HCN}(g)\) can be obtained from the reaction of \(20.0 \mathrm{~L} \mathrm{CH}_{4}(g), 20.0 \mathrm{~L} \mathrm{NH}_{3}(g)\), and \(20.0 \mathrm{~L} \mathrm{O}_{2}(\mathrm{~g})\) ? The volumes of all gases are measured at the same temperature and pressure.

Short answer

a. The balanced chemical equation for the reaction is: 2 CH4(g) + 2 NH3(g) + 3 O2(g) → 4 HCN(g) + 6 H2O(g) b. The volume of hydrogen cyanide that can be obtained from the reaction of 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g) is approximately 26.67 L.

Step-by-step solution

Write the balanced chemical equation for the given reaction.

Methane (CH4), ammonia (NH3), and oxygen (O2) react to form hydrogen cyanide (HCN) and gaseous water (H2O). The unbalanced chemical equation can be written as: CH4(g) + NH3(g) + O2(g) → HCN(g) + H2O(g) Now, balance the chemical equation: 2 CH4(g) + 2 NH3(g) + 3/2 O2(g) → 2 HCN(g) + 3 H2O(g) This can be written as fractions: CH4(g) + NH3(g) + 3/4 O2(g) → HCN(g) + 3/2 H2O(g) Or if you prefer whole numbers: 2 CH4(g) + 2 NH3(g) + 3 O2(g) → 4 HCN(g) + 6 H2O(g)

Determine the limiting reactant

We are given 20.0 L of each reactant (CH4, NH3, and O2). To find the limiting reactant, we must find the mole ratio between the reactants and compare them with the balanced chemical equation. Since the volumes of all gases are measured at the same temperature and pressure, we can assume that their mole ratio remains the same, so no calculations regarding moles are necessary here. From the balanced chemical equation, the mole ratio between CH4, NH3, and O2 should be 2:2:3. Comparing the given volumes of each reactant: CH4:NH3:O2 20:20:20 Since the given ratios are not in the same proportion as the balanced chemical equation, there must be a limiting reactant. To find it, divide the volume of each reactant by the balanced equation's coefficient: For CH4: 20/2 = 10 For NH3: 20/2 = 10 For O2: 20/3 ~ 6.67 The smallest ratio is for O2, so O2 is the limiting reactant.

Calculate the volume of hydrogen cyanide produced

Now that we know O2 is the limiting reactant, we can calculate the amount of hydrogen cyanide produced. From the balanced chemical equation: 3 O2(g) -> 4 HCN(g), so the ratio between O2 and HCN is 3:4. Volume of HCN produced = (4/3) * Volume of O2 used Volume of HCN = (4/3) * 20.0 L Volume of HCN = 26.67 L So, the volume of hydrogen cyanide that can be obtained from the reaction of 20.0 L CH4(g), 20.0 L NH3(g), and 20.0 L O2(g) is approximately 26.67 L.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves the calculations relating reactants and products in a chemical reaction. It is based on the principle that in a chemical reaction, the mass of the reactants equals the mass of the products. This concept is essential for predicting the quantities of products that can be formed in a reaction when the quantities of the reactants are known.
One of the main applications of stoichiometry is to determine the proportions in which reactants combine. By using the balanced chemical equation, you can calculate the amount of each substance needed to react completely without any leftover reactants. This is especially useful in industrial settings where cost efficiency and waste minimization are important.
In this exercise, stoichiometry helps to calculate the volume of hydrogen cyanide ( HCN ) produced based on the volumes of the reactants, using their mole ratios. Since gases are involved, the volumes directly correspond to moles under constant conditions of temperature and pressure.

Balancing Chemical Equations

Balancing chemical equations is a critical skill in chemistry as it ensures that the law of conservation of mass is obeyed. This law states that matter can neither be created nor destroyed. Thus, in a chemical equation, the number of atoms for each element must be the same on both the reactant and product sides.
To balance a chemical equation, you adjust the coefficients (numbers in front of molecules) without changing the chemical formulas of the substances involved. This process can involve some trial and error and requires careful counting of each type of atom on both sides.
For example, in the given exercise, the initial chemical equation for the reaction of methane (CH₄), ammonia (NH₃), and oxygen (O₂) is unbalanced: CH₄(g) + NH₃(g) + O₂(g) → HCN(g) + H₂O(g). By balancing the equation, we ensure accurate stoichiometric calculations. The balanced equation becomes: 2 CH₄(g) + 2 NH₃(g) + 3 O₂(g) → 4 HCN(g) + 6 H₂O(g). This balanced equation reflects the most accurate proportions of reactants and products allowing for precise stoichiometric calculations.

Limiting Reactant

The concept of a limiting reactant is integral to understanding chemical reactions. It refers to the substance that is completely consumed first during a chemical reaction and thus determines the amount of product that can be formed. Once the limiting reactant is used up, no further reaction can occur, even if other reactants are in excess.
To identify the limiting reactant, one must compare the amount of product that can be formed by each reactant. This is done by dividing the amount of each reactant by its coefficient in the balanced chemical equation, and identifying which reactant gives the smallest ratio.
In the current scenario, the balanced equation is: 2 CH₄(g) + 2 NH₃(g) + 3 O₂(g) → 4 HCN(g) + 6 H₂O(g). Given 20.0 L of each reactant (CH₄, NH₃, and O₂), the smallest ratio is found for oxygen ( O₂ ), making it the limiting reactant. With O₂ as the limiting reactant, we can calculate the maximum volume of HCN produced, ensuring that other reactants are in excess and won’t limit the amount of product. This demonstrates how understanding the limiting reactant helps in efficiently planning chemical processes.