Question

Ammonia is synthesized commercially from nitrogen gas and hydrogen gas for the production of fertilizers: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ If \(100.0 \mathrm{~g}\) of nitrogen reacts completely with excess hydrogen, and \(34.0 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) are obtained, what is the percent yield of ammonia?

Short answer

The percent yield of ammonia would be around 28.02%.

Step-by-step solution

Calculate the moles of nitrogen

First we calculate the number of moles of nitrogen using its molar mass. Nitrogen's atomic mass is \(\approx 14 \, g/mol\), and given that the molecule in question is \(N_{2}\), we have \(28 \, g/mol\) for nitrogen. Using the formula \( n = m/M\), where \(m\) is the mass and \(M\) is the molar mass, we get \( n_{N_{2}} = 100g / (28 \, g/mol) = \approx 3.57 \, mol\)

Calculate the theoretical moles and mass of ammonia

From the balanced chemical equation, we know that one mole of \(N_{2}\) reacts to form two moles of \(NH_{3}\). Therefore, we have \(3.57 \, mol \times 2 = \approx 7.14 \, mol\) of \(NH_{3}\) expected. The molar mass of \(NH_{3}\) is approximately \(17 \, g/mol\), so the expected mass (theoretical yield) is \(7.14 \, mol \times 17 \, g/mol = \approx 121.38 \, g\).

Calculate the percent yield

The percent yield is calculated by the formula \(\% \text{ yield} = ( \text{Actual yield} / \text{Theoretical yield} ) \times 100\%. \) In this case, we have \( \% \text{ yield} = (34 \, g / 121.38 \, g) \times 100\% \approx 28.02 \%.\)

Key concepts

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is vital for understanding how much of each reactant is needed to produce a certain amount of product in a chemical reaction. In the context of ammonia production, we start by looking at the balanced chemical equation:
\[ \text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g) \].
This equation tells us that one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia gas. To find out how much product can be made from a given amount of reactant, we first need to use the concept of molar masses to convert the mass of our reactants into moles, as this is the measurement unit for chemical reactions. Understanding stoichiometry allows us to predict the outcomes of reactions and is key for calculating theoretical yields, which we can compare with the actual yield to determine the efficiency of a reaction in percent yield terms.

Molar Mass Calculation

The molar mass of a substance is the weight of one mole of that substance. It's a fundamental concept in chemistry because it bridges the gap between the macroscopic world (grams of a substance) and the microscopic world (number of molecules or atoms). For instance, nitrogen has an atomic mass of approximately 14 grams per mole. However, in the reaction for ammonia production, we have dinitrogen, or N₂, so we have to double the atomic mass, yielding a molar mass of 28 grams per mole for N₂.
By dividing the mass of our nitrogen by its molar mass, we determine the number of moles reacting, which is crucial for predicting how much ammonia should form if all the nitrogen reacted perfectly (the basis for theoretical yield). Similarly, knowing the molar mass of ammonia (17 grams per mole) is essential for converting theoretical moles of ammonia into grams, which can then be compared to the actual mass obtained in an experiment.

Theoretical and Actual Yield

When we talk about yields in chemistry, we differentiate between two types: theoretical yield and actual yield. The theoretical yield is the amount of product that can be produced in a perfectly efficient reaction with no losses, based on stoichiometry. It's a calculated value that tells us what we should expect under ideal conditions. For ammonia, we determine this by using the stoichiometry from the balanced equation along with the moles of nitrogen we have.
In contrast, the actual yield is what is really produced in a laboratory or industrial setting, and it's often less due to losses or inefficiencies. When calculating the percent yield, we are essentially evaluating the efficiency of the chemical process by comparing the actual yield (34 grams of ammonia in our example) to the theoretical yield (approximately 121.38 grams of ammonia). This comparison is pivotal for identifying areas of improvement in a chemical process, aiming to get the actual yield as close to the theoretical yield as possible, signifying a more efficient and cost-effective reaction.