Question

A \(2.747-\mathrm{g}\) sample of manganese metal is reacted with excess \(\mathrm{HCl}\) gas to produce \(3.22 \mathrm{~L} \mathrm{H}_{2}(g)\) at \(373 \mathrm{~K}\) and \(0.951 \mathrm{~atm}\) and a manganese chloride compound \(\left(\mathrm{MnCl}_{x}\right)\). What is the formula of the manganese chloride compound produced in the reaction?

Short answer

The formula of the manganese chloride compound produced in the reaction is MnCl₅.

Step-by-step solution

Find the amount of H₂ gas produced

Since we're given the volume, temperature, and pressure of hydrogen gas, we can use the Ideal Gas Law to find the amount of hydrogen gas in moles. The formula for the Ideal Gas Law is: \[PV = nRT\] Where P is pressure, V is volume, n is the number of moles of gas, R is the Ideal Gas Constant (0.08206 L·atm/mol·K), and T is temperature in Kelvin. We are given the pressure (P = 0.951 atm), volume (V = 3.22 L), and temperature (T = 373 K), so we can solve for n: \[n (H_2) = \frac{PV}{RT}\]

Calculate the amount of H₂ gas in moles

Using the given values, we can calculate the value of n: \[n (H_2) = \frac{(0.951 \mathrm{~atm})(3.22 \mathrm{~L})}{(0.08206 \mathrm{~L \cdot atm/mol \cdot K})(373 \mathrm{~K})}\] After calculating, we get: \[n (H_2) = 0.124 \mathrm{~mol}\] Thus, there are 0.124 moles of H₂ gas produced.

Determine the stoichiometry of the reaction

Now, we'll use stoichiometry to determine the ratio of hydrogen gas to manganese metal. The balanced equation for the reaction is: \[\mathrm{Mn(s)} + x \mathrm{HCl(g)} \rightarrow \mathrm{MnCl}_x\mathrm{(s)} + \frac{x}{2} \mathrm{H_2(g)}\] Since we know the moles of hydrogen gas, we can determine the moles of manganese: \[0.124 \mathrm{~mol}\, H_2 \times \frac{1 \,\mathrm{mol}\, Mn}{\frac{x}{2} \, \mathrm{mol} \, H_2} = \frac{0.124 \, \mathrm{mol}}{\frac{x}{2}} \, \mathrm{mol}\, Mn\]

Calculate the moles of manganese

We are given that 2.747 g of manganese metal is reacted. To find the moles of manganese, we'll use the molar mass of manganese (54.94 g/mol): \[ \mathrm{moles\; of\; Mn} = \frac{2.747\; \mathrm{g}}{54.94\; \mathrm{g/mol}} \] After calculating, we get: \[ \mathrm{moles\; of\; Mn} = 0.0500\; \mathrm{mol} \]

Use the stoichiometry to find x

Now, we can use the relationship between the moles of manganese and hydrogen gas to find the value of x: \[\frac{0.124 \, \mathrm{mol}}{\frac{x}{2}} = 0.0500 \, \mathrm{mol}\, Mn\] Solving for x: \[x = \frac{2(0.124 \, \mathrm{mol})}{0.0500 \, \mathrm{mol}}\] After calculating, we get: \[x = 4.96 \] Since x should be an integer value, we can round x to the nearest whole number, in this case, 5.

Write the formula of the manganese chloride compound

Now that we have the value of x, we can write the formula for the manganese chloride compound: \[\mathrm{MnCl}_x = \mathrm{MnCl}_5\] So, the manganese chloride compound produced in the reaction has the formula MnCl₅.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry and physics that relates the pressure (P), volume (V), temperature (T), and the amount in moles (n) of an ideal gas. It is typically expressed as \(PV = nRT\) where R is the Ideal Gas Constant (0.08206 L·atm/mol·K).

To apply this law to real-world problems, like finding the amount of H₂ gas produced in our exercise, one needs to ensure that all the variables are in the correct units. For instance, pressure in atmospheres (atm), volume in liters (L), and temperature in Kelvin (K). This law becomes crucial when we need to find out either the amount of gas (in moles) present, as shown in the exercise, or other properties like how much space the gas will occupy under different conditions.

It's essential when working with the Ideal Gas Law to remember that it assumes ideal behavior which means the gas particles have no volume and do not interact with each other; although this isn’t completely true for real gases, it’s a close approximation for many conditions. When students tackle exercises involving gases, understanding and correctly applying the Ideal Gas Law is often the key to finding the solution.

Molar Mass

Molar mass is the mass of one mole of a substance and it is one of the most important concepts in stoichiometry. The molar mass can be found on the periodic table, as it is the atomic weight of an element expressed in grams per mole (g/mol). In the context of our exercise, the molar mass of manganese is 54.94 g/mol.

Calculating the moles of manganese requires dividing the given mass of manganese by its molar mass. This provides us with the stoichiometric 'currency' that allows us to relate masses of different compounds and elements in a chemical reaction. Understanding molar mass and how to apply it is fundamental in determining the proportions of reactants and products in chemical reactions.

Molar mass also helps in determining the percentage composition of compounds and can be crucial for figuring out empirical and molecular formulas of unknown substances. Remember, when completing exercises involving molar mass, always check that your units are consistent, and the mass of the element or compound is converted to moles correctly.

Chemical Reaction Balancing

Balancing chemical reactions is a key step in the process of stoichiometry, which involves the quantitative relationships of the substances involved in chemical reactions. For a chemical reaction to obey the law of conservation of mass, the same number of atoms of each element must exist on both the reactant and product sides of the equation.

In the exercise at hand, the key to finding the formula of the manganese chloride compound lies in balancing the manganese and hydrogen gas as evidenced by the stoichiometry of the reaction. Through balancing, we are able to determine the coefficient 'x', which represents the number of hydrogen chloride (HCl) molecules reacting with manganese (Mn) to produce manganese chloride (MnCl_x) and hydrogen gas (H₂).

Understanding how to balance chemical equations is essential for students as it is not only important for theoretical chemistry but also for practical applications such as determining the amounts of reactants or products in a given reaction.