Question

Hydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: $$2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g)+2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ If \(5.00 \times 10^{3} \mathrm{kg}\) each of \(\mathrm{NH}_{3}, \mathrm{O}_{2},\) and \(\mathrm{CH}_{4}\) are reacted, what mass of HCN and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Short answer

Based on a 100% yield, \(2.81 \times 10^3 \, kg\) of HCN and \(5.62 \times 10^3 \, kg\) of H2O will be produced from the given amounts of reactants.

Step-by-step solution

Calculate moles of reactants

Using the molar mass, we will calculate the number of moles of each reactant available: NH3: \(1 NH_3 = 14.01 (N) + 3 \cdot 1.01 (H) = 17.03 \, g/mol\) O2: \(1 O_2 = 2 \cdot 16.00 (O) = 32.00 \, g/mol\) CH4: \(1 CH_4 = 12.01 (C) + 4 \cdot 1.01 (H) = 16.04 \, g/mol \) Now, we will convert the given mass of each reactant to moles: NH3: \(\frac{5.00 \times 10^3 \, kg}{17.03 \, g/mol} = 2.94 \times 10^5 \, moles\) O2: \(\frac{5.00 \times 10^3 \, kg}{32.00 \, g/mol} = 1.56 \times 10^5 \, moles\) CH4: \(\frac{5.00 \times 10^3 \, kg}{16.04 \, g/mol} = 3.12 \times 10^5 \, moles\)

Determine the limiting reactant

Now we need to determine the limiting reactant, the reactant that will be completely used up in the reaction, by comparing the moles of each reactant with the stoichiometry given by the balanced chemical equation. We do this by finding the reactant that has the lowest ratio of moles available to the stoichiometric coefficients: NH3: \(\frac{2.94 \times 10^5}{2} = 1.47 \times 10^5\) O2: \(\frac{1.56 \times 10^5}{3} = 5.20 \times 10^4\) CH4: \(\frac{3.12 \times 10^5}{2} = 1.56 \times 10^5\) From the calculations above, O2 has the lowest ratio, which means it is the limiting reactant.

Calculate mass of HCN and H2O formed

With O2 as the limiting reactant, we can now use stoichiometry to determine the moles of HCN and H2O produced: Moles of HCN produced: \(1.56 \times 10^5 \, moles \, O_2 \cdot \frac{2 \, moles \, HCN}{3 \, moles \, O_2} = 1.04 \times 10^5 \, moles\) Moles of H2O produced: \(1.56 \times 10^5 \, moles \, O_2 \cdot \frac{6 \, moles \, H_2O}{3 \, moles \, O_2} = 3.12 \times 10^5 \, moles\) Now, let's calculate the mass of each product using their molar masses: HCN: \(1.04 \times 10^5 \, moles \cdot 27.03 \, g/mol = 2.81 \times 10^6 \, g = 2.81 \times 10^3 \, kg\) (The molar mass of HCN is calculated as \(\, 1C + 1N + 1H = 12.01 + 14.01 + 1.01 = 27.03 \, g/mol \) ) H2O: \(3.12 \times 10^5 \, moles \cdot 18.02 \, g/mol = 5.62 \times 10^6 \, g = 5.62 \times 10^3 \, kg\) (The molar mass of H2O is calculated as \(\, 2H + 1O = 2 \cdot 1.01 + 16.00 = 18.02 \, g/mol \) ) Therefore, based on a 100% yield, 2.81 x 10^3 kg of HCN and 5.62 x 10^3 kg of H2O will be produced from the given amounts of reactants.

Key concepts

Stoichiometry

Understanding stoichiometry is like learning the language of chemistry. It is a fascinating mathematical technique used to calculate the amounts of reactants and products involved in a chemical reaction. In every chemical reaction, elements and compounds interact in fixed ratios, known as stoichiometric ratios. These ratios are derived from the balanced chemical equation. A balanced equation shows the relationship between the reactants and products, allowing chemists to predict how much product will form and how much of each reactant is needed or will be used up.

In our exercise, we are given a balanced chemical equation involving ammonia (\(\text{NH}_3\)), oxygen (\(\text{O}_2\)), and methane (\(\text{CH}_4\)) as reactants, yielding hydrogen cyanide (\(\text{HCN}\)) and water (\(\text{H}_2\text{O}\)) as products. Stoichiometry helps us determine how much HCN and H2O can be formed from given masses of reactants. It is crucial to know the balanced chemical equation as it provides the stoichiometric coefficients, which are used as conversion factors in calculations to relate the amounts of reactants and products.

• Stoichiometric coefficients indicate the proportion of molecules or moles involved in the reaction.
• Using these coefficients, we calculate the number of moles for each substance based on given quantities, enabling us to identify the limiting reactant and calculate the mass of products formed.

By using stoichiometry, we can efficiently manage resources and predict outcomes in chemical manufacturing processes.

Molar Mass

The concept of molar mass plays a key role in stoichiometry. Molar mass is defined as the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is derived from the atomic masses of the elements found on the periodic table. But why is molar mass so important?

Knowing the molar mass of a compound allows us to convert between the mass of a substance and the amount in moles. Since chemical reactions involve precise amounts of substances, molar mass serves as a bridge between the scales of mass and moles. In the exercise, we computed the molar mass for ammonia, oxygen, and methane to determine how many moles of each reactant were present in 5,000 kilograms.

• Molar mass of ammonia (\(\text{NH}_3\)): calculated as 17.03 g/mol.
• Molar mass of oxygen (\(\text{O}_2\)): calculated as 32.00 g/mol.
• Molar mass of methane (\(\text{CH}_4\)): calculated as 16.04 g/mol.
• Molar mass of HCN product: calculated as 27.03 g/mol.
• Molar mass of water (\(\text{H}_2\text{O}\)): calculated as 18.02 g/mol.

This conversion allows us to easily relate the empirical quantities provided to the mole ratio dictated by the chemical equation.

Chemical Equation

A chemical equation is essentially a symbolic representation of a chemical reaction. It succinctly conveys which substances participate in the reaction, and what quantities are involved. A properly balanced chemical equation is fundamental to chemical analysis and calculations.

The given chemical equation for producing hydrogen cyanide is:\[2 \mathrm{NH}_{3}(g) + 3 \mathrm{O}_{2}(g) + 2 \mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{HCN}(g) + 6 \mathrm{H}_{2} \mathrm{O}(g)\]
Here is how to interpret different parts of the equation:

• The symbols \(\mathrm{NH}_3\), \(\mathrm{O}_2\), \(\mathrm{CH}_4\), \(\mathrm{HCN}\), and \(\mathrm{H}_2 \mathrm{O}\) represent chemical formulas of the reactants and products.
• Numbers in front of these symbols (stoichiometric coefficients) indicate the number of molecules or moles involved. For example, two moles of ammonia react with three moles of oxygen.
• The arrow (\(\longrightarrow\)) signifies the direction of the reaction, pointing from reactants to products.

By using a balanced equation, we ensure that the law of conservation of mass holds true - no atoms are lost, and the mass of reactants equals the mass of products. This balance is critical for understanding and predicting the outcomes of chemical reactions.

Chemical Reaction

At its core, a chemical reaction is a process where substances interact to form new substances. It is a transformation governed by the principles mentioned earlier, like stoichiometry and molar mass. Chemical reactions are an integral part of chemical engineering, biological processes, and everyday life.

In the exercise described, the reaction involves ammonia, oxygen, and methane combining to form hydrogen cyanide and water. Importantly, the concept of a "limiting reactant" arises – it is the reactant that will be consumed first, thus limiting the extent of the reaction. Identifying the limiting reactant is crucial because it determines how much product can be formed. In this case, oxygen was found to be the limiting reactant, meaning it was used up first and determined the maximum amount of HCN and H2O produced.

• Reactants are the starting substances: \(\mathrm{NH}_3\), \(\mathrm{O}_2\), \(\mathrm{CH}_4\)
• Products are the resulting substances: \(\mathrm{HCN}\), \(\mathrm{H}_2\mathrm{O}\)
• Limiting reactant dictates the maximum yield of products.

Understanding this concept is crucial for optimizing chemical processes and efficient resource management.