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 \(\mathrm{HCN}\) and of \(\mathrm{H}_{2} \mathrm{O}\) will be produced, assuming \(100 \%\) yield?

Short answer

Assuming a 100% yield, the reaction of 5000 kg each of ammonia, oxygen, and methane will produce approximately \(2813.3 \: \text{kg}\) of hydrogen cyanide and \(5631.3 \: \text{kg}\) of water.

Step-by-step solution

1. Convert the mass of each reactant to moles

We need to convert the mass of each reactant into moles using their respective molar masses. The molar masses of each reactant are: NH3 = 14.01 g/mol (nitrogen) + 3 * 1.01 g/mol (hydrogen) = 17.03 g/mol O2 = 2 * 16.00 g/mol (oxygen) = 32.00 g/mol CH4 = 12.01 g/mol (carbon) + 4 * 1.01 g/mol (hydrogen) = 16.04 g/mol Now convert mass to moles: moles of NH3 = (5.00*10^3 kg * 1000 g/kg) / 17.03 g/mol ≈ 293612 moles moles of O2 = (5.00*10^3 kg * 1000 g/kg) / 32.00 g/mol ≈ 156250 moles moles of CH4 = (5.00*10^3 kg * 1000 g/kg) / 16.04 g/mol ≈ 311838 moles

2. Determine the limiting reactant

To determine the limiting reactant, we need to compare the mole ratio of each reactant based on the balanced equation. The mole ratio is 2:3:2 (NH3 : O2 : CH4). So we'll divide each reactant's moles by their respective coefficients: NH3: 293612 moles / 2 ≈ 146806 O2: 156250 moles / 3 ≈ 52083 CH4: 311838 moles / 2 ≈ 155919 The lowest value belongs to O2. Therefore, oxygen (O2) is the limiting reactant.

3. Use stoichiometry to calculate the moles of products formed

From the balanced equation, we know that the mole ratio of O2 : HCN : H2O is 3:2:6. Use stoichiometry to calculate the moles of HCN and H2O formed from the moles of the limiting reactant (O2): moles of HCN = (2 moles HCN / 3 moles O2) * 156250 moles O2 ≈ 104167 moles HCN moles of H2O = (6 moles H2O / 3 moles O2) * 156250 moles O2 ≈ 312500 moles H2O

4. Convert the moles of products to mass

Now we need to convert the moles of each product to mass. To do this, we'll multiply the moles of each product by their respective molar masses: Molar masses of products: HCN = 12.01 g/mol (carbon) + 1.01 g/mol (hydrogen) + 14.01 g/mol (nitrogen) = 27.03 g/mol H2O = 2 * 1.01 g/mol (hydrogen) + 16.00 g/mol (oxygen) = 18.02 g/mol Mass of HCN = 104167 moles * 27.03 g/mol ≈ 2813292 g or 2813.3 kg Mass of H2O = 312500 moles * 18.02 g/mol ≈ 5631250 g or 5631.3 kg

Conclusion

Assuming a 100% yield, the reaction of 5000 kg each of ammonia, oxygen, and methane will produce approximately 2813.3 kg of hydrogen cyanide and 5631.3 kg of water.

Key concepts

Limiting Reactant

Understanding the concept of a limiting reactant is crucial when trying to predict the amounts of products formed in a chemical reaction. Imagine you are baking cookies and have a certain amount of flour and sugar. If you run out of flour first, it doesn't matter how much sugar you have left; you can't make more cookies. Similarly, in a chemical reaction, the limiting reactant is the substance that is completely consumed first and, as a result, determines the maximum amount of product that can be formed.

To identify the limiting reactant, you first need to convert the amounts of reactants to moles, usually using molar masses. Then, compare the mole ratios from the balanced chemical equation to the moles of each reactant available. The reactant that has the lowest ratio (when comparing actual moles to needed moles) is the limiting reactant. In the example problem given, oxygen (O₂) was identified as the limiting reactant by comparing the ratios of moles of reactants to their stoichiometric coefficients in the balanced equation.

Mole-to-Mass Conversion

The mole-to-mass conversion is a fundamental concept in chemistry that allows us to relate quantities of substances in reactions to measurable amounts. Each substance has a unique molar mass (the mass of one mole of a substance) that serves as a conversion factor between the number of moles and the mass.

To carry out this conversion, you simply multiply the number of moles of the substance by its molar mass. The molar mass is expressed in grams per mole (g/mol), which facilitates the conversion from moles to grams, and then, if necessary, to kilograms. This step is essential for translating the theoretical aspects of a chemical equation into practical, real-world quantities, such as predicting the mass of products formed from a given amount of reactants as seen in the exercise provided.

Chemical Yield Calculation

After determining the products from the limiting reactant using stoichiometry, the final step is calculating the yield—or how much product is actually produced. The theoretical yield is the maximum possible amount of product calculated through stoichiometry, assuming everything reacts perfectly without any losses.

However, real chemical reactions may deviate, leading to an actual yield that is often less than the theoretical yield. The percent yield is a measure of the efficiency of a reaction, calculated by dividing the actual yield by the theoretical yield and multiplying by 100%. In the problem we are discussing, it is assumed that the reaction has a 100% yield, which means all of the limiting reactant was converted into product, and the calculated mass represents the actual amount of products formed.