Question

A 2.747 -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 atm and a manganese chloride compound \(\left(\mathrm{Mn} \mathrm{Cl}_{x}\right)\). 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

Determine the number of moles of manganese metal and hydrogen gas

We have the mass of manganese (2.747g) and can calculate the number of moles of manganese with the help of its molar mass (54.94g/mol): \(moles_{Mn} = \frac{mass_{Mn}}{Molar \: mass} = \frac{2.747g}{54.94g/mol} = 0.0500 mol\) We have the volume, temperature, and pressure of the hydrogen gas produced, and can use the Ideal Gas Law (PV=nRT) to find the moles of hydrogen gas: \(P = 0.951 \: atm\) \(V = 3.22 \: L\) \(T = 373 \: K\) \(R = 0.0821\: \frac{L \cdot atm}{mol \cdot K}\) \(n_{H_2} = \frac{PV}{RT} = \frac{(0.951atm)(3.22L)}{(0.0821\frac{L\cdot atm}{mol \cdot K})(373K)} = 0.123\: mol\)

Use stoichiometry to find the mole ratio between manganese and chlorine

We know the balanced reaction should look like: \(Mn + xHCl \rightarrow MnCl_x + \frac{x}{2}H_2\) From Step 1, we found the moles of Mn and H2 produced: \(0.0500 \: mol \: Mn\) \(0.123 \: mol \: H_2\) Since the mole ratio between Mn and H2 in the balanced reaction is 1: \(\frac{x}{2}\), we have: \(\frac{0.0500}{1} = \frac{0.123}{\frac{x}{2}} \Rightarrow x = 4.94\) Now, since x must be a whole number, we round it to the nearest integer, which is 5.

Determine the formula of the manganese chloride compound

Now that we know the value of x, we can determine the formula of the MnClx compound: The formula of the manganese chloride compound is MnCl5.

Key concepts

Stoichiometry

Stoichiometry is a fundamental concept in chemistry that involves calculating the amounts of reactants and products in a chemical reaction. This is based on the balanced chemical equation. In this exercise, stoichiometry helps us understand how manganese reacts with hydrochloric acid to produce manganese chloride and hydrogen gas.

The balanced chemical equation for the reaction is:

• \( \mathrm{Mn} + x \mathrm{HCl} \rightarrow \mathrm{MnCl}_x + \frac{x}{2} \mathrm{H}_2 \)

Using stoichiometry, we calculate the mole ratio between manganese and hydrogen gas produced. We find the moles of manganese and hydrogen using their respective data - mass for manganese and conditions for hydrogen using the Ideal Gas Law.

Understanding stoichiometry is crucial as it enables you to work out quantitative relationships from the qualitative information in a balanced equation. As seen in this exercise, determining the amount of a product formed requires determining mole ratios. These ratios are then used to derive the formula of the unknown manganese chloride compound.

Ideal Gas Law

The Ideal Gas Law is an essential principle taught in chemistry, represented by the formula:

• \( PV = nRT \)

This law describes the relationship between pressure \( P \), volume \( V \), and temperature \( T \) of a gas with its number of moles \( n \) and the ideal gas constant \( R \). In this problem, you are applying the Ideal Gas Law to find out how many moles of hydrogen gas \( H_2 \) are produced.

Given that the pressure \( P \) is 0.951 atm, the volume \( V \) is 3.22 L, and the temperature \( T \) is 373 K, you can rearrange the Ideal Gas Law to solve for the number of moles \( n \):

• \( n_{H_2} = \frac{PV}{RT} = \frac{(0.951 \, \mathrm{atm})(3.22 \, \mathrm{L})}{(0.0821 \, \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}})(373 \, \mathrm{K})} = 0.123 \, \mathrm{mol} \)

Understanding how to use the Ideal Gas Law is vital because it allows you to link physical gas conditions to the chemical quantities in a reaction. This connection illustrates how theoretical chemistry principles apply to real-world situations.

Chemical Formulas

Chemical formulas represent the elements present in a compound and the ratios in which they combine. For example, in the exercise, we are tasked with determining the formula of the manganese chloride compound formed.

The formula \( \mathrm{MnCl}_x \) indicates that manganese is combined with a certain number \( x \) of chlorine atoms. Determining this value involves understanding the stoichiometric relationship between manganese and hydrogen as illustrated above. The calculated value of \( x = 4.94 \) is rounded to 5, giving us \( \mathrm{MnCl}_5 \).

This formula communicates that each manganese atom is accompanied by five chlorine atoms, providing insights into the compound's structure and reactivity. Learning to decipher and work with chemical formulas is crucial, as these formulas allow chemists to predict how substances will interact in chemical reactions. It helps in comprehensively understanding the identity and properties of chemical compounds.