Question

Hydrogen cyanide is prepared commercially by the reaction of methane, ${\mathrm{CH}}_4(g),$ ammonia, ${\mathrm{NH}}_3(g),$ and oxygen, ${\mathrm{O}}_2(g),$ 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 $\mathrm{L}{\mathrm{O}}_2(g)?$ The volumes of all gases are measured at the same temperature and pressure.

Short answer

The balanced chemical equation for the reaction is $2{\mathrm{CH}}_4(g)+6{\mathrm{NH}}_3(g)+3{\mathrm{O}}_2(g)\to 6\mathrm{HCN}(g)+6{\mathrm{H}}_2\mathrm{O}(g)$. The limiting reactant is ${\mathrm{NH}}_3(g)$, and the volume of $\mathrm{HCN}(g)$ obtained from the reaction is $6.67\mathrm{L}$.

Step-by-step solution

Write and balance the chemical equation for the reaction

The given information indicates that ${\mathrm{CH}}_4(g),$ ${\mathrm{NH}}_3(g),$ and ${\mathrm{O}}_2(g)$ react to produce hydrogen cyanide, $\mathrm{HCN}(g),$ and gaseous water, ${\mathrm{H}}_2\mathrm{O}(g).$ First, write the unbalanced equation and then balance it with the appropriate coefficients. Unbalanced equation: $${\mathrm{CH}}_4(g)+{\mathrm{NH}}_3(g)+{\mathrm{O}}_2(g)\to \mathrm{HCN}(g)+{\mathrm{H}}_2\mathrm{O}(g)$$ Now balance the equation: $$2{\mathrm{CH}}_4(g)+6{\mathrm{NH}}_3(g)+3{\mathrm{O}}_2(g)\to 6\mathrm{HCN}(g)+6{\mathrm{H}}_2\mathrm{O}(g)$$

Determine the limiting reactant

Since all of the volumes of gases are given and are measured at the same temperature and pressure, we can use the balanced chemical equation to determine the ratio between the reactants. The limiting reactant is the one that is used up first, as it determines the amount of product that can be formed. From the balanced equation, we have: $$2{\mathrm{CH}}_4(g):6{\mathrm{NH}}_3(g):3{\mathrm{O}}_2(g)\Rightarrow 1{\mathrm{CH}}_4(g):3{\mathrm{NH}}_3(g):1\frac{1}{2}{\mathrm{O}}_2(g)$$ Comparing the molar ratios of the reactants, we have: $$\frac{20.0{\mathrm{L}\mathrm{CH}}_4(g)}{1}:\frac{20.0{\mathrm{L}\mathrm{NH}}_3(g)}{3}:\frac{20.0{\mathrm{L}\mathrm{O}}_2(g)}{\frac{3}{2}}\Rightarrow 20.0:6.67:13.33$$ Since the ratio of ${\mathrm{NH}}_3$ is the smallest number, ${\mathrm{NH}}_3$ is the limiting reactant.

Calculate the volume of HCN produced

As we found that the limiting reactant is ${\mathrm{NH}}_3,$ we will now use the stoichiometric coefficients to calculate the volume of $\mathrm{HCN}(g)$ produced. From the balanced chemical equation, we can see that $6\mathrm{moles}$ of ${\mathrm{NH}}_3$ produce $6\mathrm{moles}$ of $\mathrm{HCN}.$ Therefore, the ratio is $1:1,$ and the volume of $\mathrm{HCN}$ produced will be equal to the volume of the limiting reactant, ${\mathrm{NH}}_3$ used. Since we are given $20.0\mathrm{L}$ of ${\mathrm{NH}}_3,$ the volume of $\mathrm{HCN}$ produced is $\frac{20.0\mathrm{L}}{3}=6.67\mathrm{L}$ of $\mathrm{HCN}(g)$. So, the volume of $\mathrm{HCN}(g)$ obtained from the reaction of given volumes of ${\mathrm{CH}}_4(g),$ ${\mathrm{NH}}_3(g),$ and ${\mathrm{O}}_2(g)$ is $6.67\mathrm{L}$.

Key concepts

Chemical equation balancing

In chemistry, equations describe how substances react with each other. However, these equations must be balanced, ensuring that atoms of each element are conserved during the reaction. This is rooted in the law of conservation of mass, which maintains that mass in a closed system must remain constant. To balance a chemical equation, you adjust the coefficients (the numbers placed before compounds) to ensure that the number of atoms for each element is equal on both the reactant and product sides.

For the equation given in the exercise, we start with an unbalanced version: $${\mathrm{CH}}_4(g)+{\mathrm{NH}}_3(g)+{\mathrm{O}}_2(g)\to \mathrm{HCN}(g)+{\mathrm{H}}_2\mathrm{O}(g)$$ Balancing this requires ensuring equal numbers of each element type are present on both sides. The balanced equation becomes:$$2{\mathrm{CH}}_4(g)+6{\mathrm{NH}}_3(g)+3{\mathrm{O}}_2(g)\to 6\mathrm{HCN}(g)+6{\mathrm{H}}_2\mathrm{O}(g)$$ Here, every element (C, H, N, O) has the same quantity on both sides of the equation, satisfying the law of conservation of mass. This method of balancing ensures that the equation accurately represents the reaction, both qualitatively and quantitatively.

Stoichiometry

Stoichiometry is the study of quantitative relationships in chemical reactions. It allows chemists to predict how much of each substance is needed or produced in a reaction. By analyzing the coefficients from balanced chemical equations, we can derive ratios that tell us about these relationships among reactants and products.

In the problem, after balancing the equation, the stoichiometric coefficients are found to be $2$ for ${\mathrm{CH}}_4$, $6$ for ${\mathrm{NH}}_3$, and $3$ for ${\mathrm{O}}_2$. The same coefficients apply to the products: $6$ for $\mathrm{HCN}$ and $6$ for ${\mathrm{H}}_2\mathrm{O}$.
The Stoichiometric Ratios:

• 2 moles of ${\mathrm{CH}}_4$ reacts with 6 moles of ${\mathrm{NH}}_3$ and 3 moles of ${\mathrm{O}}_2$ to produce 6 moles of $\mathrm{HCN}$.

This same numerical relationship can be applied directly to the gas volumes (at constant temperature and pressure) in the exercise, leveraging Avogadro's law that states equal volumes of gases at the same temperature and pressure contain an equal number of molecules.

Gas volumes in chemical reactions

When dealing with gases in chemical reactions, it's important to understand how volume changes can be connected through stoichiometry, especially under conditions of constant temperature and pressure. Thanks to Avogadro’s Law, you can treat gas volumes as directly proportional to the number of moles.

In the given exercise, the initial volumes of all reactant gases are equal ($20.0\mathrm{L}$). By applying the stoichiometric coefficients from the balanced equation, we can determine how much of each gas volume turns into product.
The Steps:

• Start by identifying the potential limiting reactant through comparison of initial ratios to stoichiometrical ratios.
• For this exercise, ${\mathrm{NH}}_3$ is the limiting reactant because it runs out first due to the higher consumption ratio (6:6 compared to 2:6 and 3:6 for ${\mathrm{CH}}_4$ and ${\mathrm{O}}_2$, respectively).
• As a result, the volume of the product formed ($\mathrm{HCN}$) is directly related to the volume of ${\mathrm{NH}}_3$, leading to a calculated product volume of $6.67\mathrm{L}$.

This highlights the principle that under consistent conditions, gas volumes in reactions can be calculated simplistically when stoichiometry and gas laws synergize. It’s an easy and effective way to predict outcomes in gaseous reactions.