Unlocking Alkane Structure: A Chlorination Chemistry Mystery

Hey chemistry enthusiasts! Today, we're diving deep into an intriguing problem involving the chlorination of an unknown alkane. We'll break down the steps, unravel the reactions, and ultimately deduce the alkane's structure. So, buckle up and let's get started!

The Chlorination Conundrum: A Step-by-Step Breakdown

In this chlorination puzzle, we're presented with a gaseous alkane of unknown structure that reacts with chlorine under diffused light. The key here is that the reaction leads to complete chlorination, meaning all hydrogen atoms in the alkane are replaced by chlorine atoms. We also have a crucial piece of information: the volume of hydrogen chloride (HCl) produced is six times the initial volume of the alkane. To kick things off, the initial volume of the mixture is given as 44 liters. Let's dissect this and solve it like true chemists!

Understanding Alkane Chlorination

Before we jump into the calculations, let's quickly recap what happens during alkane chlorination. Alkanes are saturated hydrocarbons, meaning they contain only single bonds between carbon and hydrogen atoms. Chlorination is a substitution reaction where hydrogen atoms are replaced by chlorine atoms. This reaction typically requires ultraviolet light or heat to initiate the process. The general reaction can be represented as:

CₙH₂ₙ₊₂ + xCl₂ → CₙH₂ₙ₊₂₋ₓClₓ + xHCl

Where:

• CₙH₂ₙ₊₂ represents the alkane.
• x is the number of chlorine atoms that substitute hydrogen atoms.
• CₙH₂ₙ₊₂₋ₓClₓ is the chlorinated alkane.
• HCl is hydrogen chloride.

The stoichiometry of this reaction is crucial. For every chlorine atom that replaces a hydrogen atom, one molecule of HCl is produced. This 1:1 relationship between chlorine substitution and HCl formation is the key to solving our mystery.

Analyzing the Volume Relationship

The problem states that the volume of HCl produced is six times the initial volume of the alkane. Let's denote the volume of the alkane as V alkane. According to the problem:

V(HCl) = 6 * V(alkane)

This tells us that for every molecule of alkane, six molecules of HCl are produced. Since each chlorine atom that substitutes a hydrogen atom generates one molecule of HCl, this implies that six hydrogen atoms in the alkane molecule have been replaced by chlorine atoms. Therefore, the alkane must have at least six replaceable hydrogen atoms.

Determining the Alkane's Formula

Now, let's think about the general formula for alkanes: CₙH₂ₙ₊₂. If six hydrogen atoms are replaced, it means the alkane must have had at least six hydrogen atoms to begin with. In other words, 2n + 2 must be greater than or equal to 6.

2n + 2 ≥ 6

Solving for n:

2n ≥ 4 n ≥ 2

So, the smallest possible alkane that could satisfy this condition is ethane (C₂H₆). However, ethane has only six hydrogen atoms, and if all are replaced, we'd get C₂Cl₆ and 6HCl, which fits our condition. But let's consider other possibilities.

If n = 3, we have propane (C₃H₈), which has eight hydrogen atoms. If six are replaced, we'd get C₃H₂Cl₆ and 6HCl. This also fits our condition. What if n = 4?

If n = 4, we have butane (C₄H₁₀), which has ten hydrogen atoms. Replacing six gives us C₄H₄Cl₆ and 6HCl. So, butane could also be a contender. But how do we narrow it down?

Using the Initial Volume of the Mixture

We know the initial volume of the mixture (alkane + chlorine) is 44 liters. We also know that the reaction goes to completion, meaning all the alkane reacts. Let's denote the volume of chlorine as V(Cl₂). Then:

V(alkane) + V(Cl₂) = 44 L

To determine which alkane it is, we need to relate the volume of chlorine to the volume of the alkane based on the stoichiometry of the chlorination reaction. Let's revisit the general equation:

CₙH₂ₙ₊₂ + xCl₂ → CₙH₂ₙ₊₂₋ₓClₓ + xHCl

We know x = 6 (since six HCl molecules are produced). So, for every molecule of alkane, six molecules of chlorine are consumed. In terms of volume, this means:

V(Cl₂) = 6 * V(alkane)

Now we have two equations:

1. V(alkane) + V(Cl₂) = 44 L
2. V(Cl₂) = 6 * V(alkane)

Substituting equation (2) into equation (1):

V(alkane) + 6 * V(alkane) = 44 L

7 * V(alkane) = 44 L

V(alkane) = 44 L / 7 ≈ 6.29 L

Now we know the volume of the alkane is approximately 6.29 liters. Let's calculate the volume of chlorine:

V(Cl₂) = 6 * 6.29 L ≈ 37.74 L

Identifying the Alkane: Putting It All Together

So, we've established that we have approximately 6.29 liters of alkane and 37.74 liters of chlorine. The fact that six chlorine molecules react with each alkane molecule is a big clue. Let's look back at our candidates: ethane, propane, and butane.

• Ethane (C₂H₆): All six hydrogen atoms can be replaced. The reaction would be C₂H₆ + 6Cl₂ → C₂Cl₆ + 6HCl.
• Propane (C₃H₈): Only six hydrogen atoms are replaced. The reaction would be C₃H₈ + 6Cl₂ → C₃H₂Cl₆ + 6HCl. Two hydrogen atoms remain.
• Butane (C₄H₁₀): Only six hydrogen atoms are replaced. The reaction would be C₄H₁₀ + 6Cl₂ → C₄H₄Cl₆ + 6HCl. Four hydrogen atoms remain.

The key here is that the problem specifies complete chlorination, implying that as many hydrogen atoms as possible are replaced, but not necessarily all. If we consider the fact that six HCl molecules are produced, this means six chlorine atoms have substituted six hydrogen atoms. This aligns perfectly with our calculations.

Given the volume ratios and the production of 6 moles of HCl per mole of alkane, ethane (C₂H₆) fits the criteria perfectly. All six hydrogen atoms are substituted, and the stoichiometry aligns with the volume ratios calculated.

Final Answer: The Alkane Revealed

After meticulously analyzing the reaction conditions, volume relationships, and stoichiometry, we've successfully identified the alkane! The unknown alkane is ethane (C₂H₆). The complete chlorination reaction is:

C₂H₆ + 6Cl₂ → C₂Cl₆ + 6HCl

Isn't it fascinating how we can piece together chemical puzzles using fundamental principles and a bit of logical deduction? Keep exploring, keep questioning, and happy chemistry-ing!

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