Question

Consider the following chemical equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) $$ What minimum total volume of reactant gases is needed to produce \(4.00 \mathrm{~g}\) of ammonia if the volume of each gas is measured at 11 ' \(\mathrm{C}\) and 0.998 atm?

Short answer

The minimum total volume of reactant gases needed to produce 4.00 g of ammonia under the given conditions of 11 °C and 0.998 atm is 10.896 L.

Step-by-step solution

Convert mass of ammonia to moles

To convert the mass of ammonia (4.00 g) to moles, we use its molar mass: Molar mass of NH₃ = 14.01 g/mol (N) + 3 × 1.01 g/mol (H) = 17.03 g/mol Now, let's calculate the number of moles in 4.00 g of NH₃: Moles of NH₃ = \( \frac{4.00}{17.03} \) = 0.235 mol

Determine the moles of N₂ and H₂ gases required

Using stoichiometry from the chemical equation, we can calculate the moles of N₂ and H₂ gases needed: N₂: 1 mol N₂ → 2 mol NH₃ H₂: 3 mol H₂ → 2 mol NH₃ So, to produce 0.235 mol NH₃, we need: Moles of N₂ = \( \frac{1}{2}\) × 0.235 mol = 0.1175 mol Moles of H₂ = \( \frac{3}{2}\) × 0.235 mol = 0.3525 mol

Calculate the volume of N₂ and H₂ using the ideal gas law

Now, we have the moles of N₂ and H₂ gases required under the given temperature (11 °C = 284.15 K) and pressure (0.998 atm). We can use the ideal gas law (PV = nRT) to calculate the volume of each gas. For N₂ gas: P = 0.998 atm n = 0.1175 mol R = 0.0821 L atm / (mol K) (ideal gas constant) T = 284.15 K V (N₂) = \( \frac{nRT}{P} \) = \(\frac{(0.1175)(0.0821)(284.15)}{0.998}\) = 2.724 L For H₂ gas: P = 0.998 atm n = 0.3525 mol R = 0.0821 L atm / (mol K) (ideal gas constant) T = 284.15 K V (H₂) = \(\frac{nRT}{P}\) =\(\frac{(0.3525)(0.0821)(284.15)}{0.998}\) = 8.172 L

Calculate the minimum total volume of reactant gases

Finally, we add the volumes of nitrogen and hydrogen gases to find the minimum total volume of reactant gases: Total Volume = V (N₂) + V (H₂) = 2.724 L + 8.172 L = 10.896 L The minimum total volume of reactant gases needed to produce 4.00 g of ammonia under the given conditions is 10.896 L.

Key concepts

Chemical Equation

Understanding a chemical equation is crucial for solving stoichiometry problems. A chemical equation represents the reactants and products in a chemical reaction using chemical formulas. For example, the equation
\[ \mathrm{N}_{2}(g) + 3\mathrm{H}_{2}(g) \rightarrow 2\mathrm{NH}_{3}(g) \]
shows the reaction where nitrogen gas (\(\mathrm{N}_{2}\)) reacts with hydrogen gas (\(\mathrm{H}_{2}\)) to form ammonia (\(\mathrm{NH}_{3}\)). In this reaction, 1 molecule of nitrogen reacts with 3 molecules of hydrogen to produce 2 molecules of ammonia. This is a balanced equation, meaning the number of atoms for each element is the same on both sides, a fundamental law known as the conservation of mass.
When interpreting the equation, pay attention to the coefficients—they tell us the mole-to-mole ratios needed to execute the reaction, which are essential for mole-to-mole conversions.

Ideal Gas Law

The ideal gas law is an equation of state for a hypothetical ideal gas. It establishes a relationship between pressure (P), volume (V), the number of moles (n), and temperature (T), expressed as
\[ PV = nRT \]
where R is the ideal gas constant. This law is extremely useful in calculating the volume of gases at a given temperature and pressure, as seen in our exercise. Remember, when using the ideal gas law, temperature must be in Kelvin. You can convert from Celsius to Kelvin by adding 273.15 to the Celsius temperature. In our example, to find the minimum total volume of reactant gases needed, we first calculated the moles of each reactant and then applied the ideal gas law to find their volumes under specific conditions.

Molar Mass

The molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It's calculated by adding the atomic masses of all atoms in a molecule. For instance, ammonia (\(\mathrm{NH}_{3}\)) has a molar mass of 17.03 g/mol, calculated by adding the atomic mass of nitrogen (14.01 g/mol) and three times that of hydrogen (3 × 1.01 g/mol). Knowing the molar mass allows us to convert between mass and moles of a substance, a vital step in solving stoichiometric problems. In the given exercise, we first converted the mass of ammonia to moles using its molar mass before proceeding with further calculations.

Mole-to-Mole Conversion

Mole-to-mole conversion is the process of using the coefficients from a balanced chemical equation to convert moles of one substance to moles of another. It's based on the ratio provided by the equation, which in the case of ammonia production is 1 mole of nitrogen to 3 moles of hydrogen producing 2 moles of ammonia. If the problem provides the moles of one substance, we can find the moles of another substance in the reaction using this ratio. For our example, we determined the moles of reactants needed to produce a certain amount of ammonia and then used the ideal gas law to find their respective volumes, all thanks to mole-to-mole conversion.