Question

Consider the following chemical equation: $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g) $$ What volumes of nitrogen gas and hydrogen gas, each measured at \(11^{\circ} \mathrm{C}\) and 0.998 atm, are needed to produce \(5.00 \mathrm{g}\) of ammonia?

Short answer

The volumes of nitrogen gas and hydrogen gas required to produce 5.00 g of ammonia at 11°C and 0.998 atm are approximately 3.44 L for nitrogen gas and 10.32 L for hydrogen gas.

Step-by-step solution

Convert grams of ammonia to moles

To find the moles of ammonia produced, we use the molar mass of ammonia, which is 17.03 g/mol. \(n_{NH_3} = \frac{5.00 \, g}{17.03 \, g/mol} = 0.2938 \, moles\)

Determine the moles of nitrogen and hydrogen required using stoichiometry

From the balanced chemical equation, 1 mole of nitrogen reacts with 3 moles of hydrogen to produce 2 moles of ammonia: \(\mathrm{N}_{2} + 3 \, \mathrm{H}_{2} \rightarrow 2 \, \mathrm{NH}_{3}\) Therefore, we can use stoichiometry to find the moles of nitrogen and hydrogen required: Moles of nitrogen = \(0.2938 \, mol \times \frac{1 \, mol \, N_2}{2 \, mol \, NH_3} = 0.1469 \, mol\) Moles of hydrogen = \(0.2938 \, mol \times \frac{3 \, mol \, H_2}{2 \, mol \, NH_3} = 0.4407 \, mol\)

Convert temperature to Kelvin

The temperature needs to be in Kelvin for the ideal gas law equation. Convert 11°C to Kelvin: \(T = 11 + 273.15 = 284.15 \, K\)

Apply the ideal gas law to find the volume of nitrogen and hydrogen

The ideal gas law equation is: \(PV = nRT\), where P is pressure, V is volume, n is moles, R is the ideal gas constant, and T is temperature. We will use the ideal gas constant with an appropriate value for the pressure unit (atm), R = 0.0821 L·atm/mol·K. For nitrogen gas, we have: \(0.998 \, atm \times V_{N_2} = 0.1469 \, mol \times 0.0821 \, \frac{L \cdot atm}{mol \cdot K} \times 284.15 \, K\) Solve for the volume of nitrogen gas, \(V_{N_2}\): \(V_{N_2} = \frac{0.1469 \, mol \times 0.0821 \, \frac{L \cdot atm}{mol \cdot K} \times 284.15 \, K}{0.998 \, atm} = 3.441 \, L\) For hydrogen gas, we have: \(0.998 \, atm \times V_{H_2} = 0.4407 \, mol \times 0.0821 \, \frac{L \cdot atm}{mol \cdot K} \times 284.15 \, K\) Solve for the volume of hydrogen gas, \(V_{H_2}\): \(V_{H_2} = \frac{0.4407 \, mol \times 0.0821 \, \frac{L \cdot atm}{mol \cdot K} \times 284.15 \, K}{0.998 \, atm} = 10.32 \, L\)

State the answer

The volumes of nitrogen gas and hydrogen gas required to produce 5.00 g of ammonia at 11°C and 0.998 atm are approximately 3.44 L for nitrogen gas and 10.32 L for hydrogen gas.

Key concepts

Ideal Gas Law

The ideal gas law is a profound equation that bridges the gap between chemistry and physics, offering insights into the behavior of gases under various conditions. It can be stated in the form of an equation: \[PV = nRT\]which describes the relationship between the pressure (\(P\)), volume (\(V\)), amount in moles of gas (\(n\)), temperature in Kelvin (\(T\)), and the ideal gas constant (\(R\)). The beauty of this equation lies in its ability to accurately predict how a gas will react when subjected to changes in any of these variables.

In the example given, the ideal gas law is crucial for determining the volume of nitrogen gas and hydrogen gas needed to produce a certain mass of ammonia. By manipulating the equation, we can solve for the volume once the other variables are known. In educational contexts, it's imperative to understand not just how to use the equation, but why it represents the behavior of gases so well. The assumption behind the ideal gas law is that it applies to 'ideal' gases, which are hypothetical gases whose molecules don't interact with each other and take up no space. Real gases can often be approximated as ideal, especially under conditions of high temperature and low pressure.

Applying the Ideal Gas Law in Stoichiometry

When intertwined with stoichiometry, the ideal gas law becomes a valuable tool to predict the outcomes of chemical reactions involving gases. Given our problem, after using stoichiometry to find the moles of gases involved, the ideal gas law allows us to find the corresponding volume. Remember, the values plugged into the equation need to be consistent, so we convert our temperature into Kelvin and ensure our units match for pressure and the gas constant.
Although using the ideal gas constant, R = 0.0821 L·atm/mol·K, may seem arbitrary, it's chosen to ensure the units remain consistent when pressure is in atmospheres (atm) and volume in liters (L). The importance of unit consistency cannot be overstated—it's essential to obtaining accurate results.

Chemical Equation Balancing

To understand the stoichiometry of a chemical reaction, one first must master the art of balancing chemical equations. At its core, this skill ensures that the principle of the conservation of mass is respected; the quantity of each element must remain constant before and after the reaction. This translates to having the same number of atoms of each element on both the reactants and products side of a chemical equation.

In our given exercise, the chemical equation for the synthesis of ammonia is already balanced: \[\text{N}_2(g) + 3 \text{H}_2(g) \rightarrow 2 \text{NH}_3(g)\]Notice that there are two nitrogen atoms and six hydrogen atoms on both sides of the equation. Balancing equations is often a precursor to stoichiometry calculations, as it determines the mole ratios that are essential for calculating reactants and products.

From Balanced Equations to Moles

Once balanced, the equation provides a ratio of how many moles of each reactant are needed to produce a certain amount of product. For the reaction at hand, it tells us one mole of nitrogen gas reacts with three moles of hydrogen gas to produce two moles of ammonia. These ratios become the foundation upon which we apply stoichiometry to deduce how much reactants are required to produce a desired quantity of a product, solving real-world problems like the one presented in this exercise.

Molar Mass

Molar mass is a fundamental characteristic of substances in chemistry, defining the mass of one mole of a substance. It essentially tells us how much one mole of any element or compound weighs, and is of paramount importance when converting between grams and moles—a frequent necessity in chemistry.

For instance, in our sample problem, we use the molar mass of ammonia, which is 17.03 g/mol, to convert from grams into moles. This conversion is the first critical step in a stoichiometry problem, bridging the gap between the mass of a compound measured often in a laboratory with the mole concept, which is central to chemical reactions.

Understanding and Using Molar Mass

Understanding molar mass allows students to navigate seamlessly across different types of measurements. It plays a pivotal role in solving stoichiometry problems because stoichiometry is based on the mole concept, not on grams or any other unit of mass. This means that when we state the quantities of substances reacting or produced, we refer to moles, making the molar mass a conversion factor that is always relevant when we start with or seek a mass quantity. When it comes to practical applications, knowing the molar mass allows chemists to measure out precise amounts of substances needed for reactions, just as we needed to calculate the quantity of nitrogen and hydrogen gases to produce a specific amount of ammonia in our exercise.