Question

A 20 ml mixture of $C_2H_4$ and $C_2H_2$ undergoes sparking in gas eudiometer with just sufficient amount of $O_2$ and shows contraction of 37.5 ml. The volume (in ml) of $C_2H_2$ in the mixture is.

Answer 1

5

Answer 2

To solve this problem, we need to use the principles of gas eudiometry and stoichiometry. We will write balanced chemical equations for the combustion of $C_2H_4$ and $C_2H_2$ and then set up a system of equations based on the given volumes and contraction.

Let $V_1$ be the volume of $C_2H_4$ in the mixture and $V_2$ be the volume of $C_2H_2$ in the mixture. Given that the total volume of the mixture is 20 ml: $V_1 + V_2 = 20$ ml (Equation 1)

1. Combustion of $C_2H_4$: The balanced chemical equation for the combustion of ethene ($C_2H_4$) is: $C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)$

According to Gay-Lussac's Law of Gaseous Volumes, the volumes of gaseous reactants and products bear a simple whole-number ratio to one another, provided the temperature and pressure are constant. Water formed is in the liquid state, so its volume is negligible in gas eudiometry experiments.

For $V_1$ ml of $C_2H_4$:

• Volume of $O_2$ consumed = $3V_1$ ml
• Volume of $CO_2$ produced = $2V_1$ ml
• Initial volume of gaseous reactants = $V_1 + 3V_1 = 4V_1$ ml
• Final volume of gaseous products = $2V_1$ ml
• Contraction due to $C_2H_4$ combustion ($C_1$) = (Initial volume of gaseous reactants) - (Final volume of gaseous products) $C_1 = 4V_1 - 2V_1 = 2V_1$ ml

2. Combustion of $C_2H_2$: The balanced chemical equation for the combustion of ethyne ($C_2H_2$) is: $C_2H_2(g) + \frac{5}{2}O_2(g) \rightarrow 2CO_2(g) + H_2O(l)$ To avoid fractions, we can write it as: $2C_2H_2(g) + 5O_2(g) \rightarrow 4CO_2(g) + 2H_2O(l)$

For $V_2$ ml of $C_2H_2$:

• Volume of $O_2$ consumed = $\frac{5}{2}V_2 = 2.5V_2$ ml
• Volume of $CO_2$ produced = $2V_2$ ml
• Initial volume of gaseous reactants = $V_2 + 2.5V_2 = 3.5V_2$ ml
• Final volume of gaseous products = $2V_2$ ml
• Contraction due to $C_2H_2$ combustion ($C_2$) = (Initial volume of gaseous reactants) - (Final volume of gaseous products) $C_2 = 3.5V_2 - 2V_2 = 1.5V_2$ ml

3. Total Contraction: The total observed contraction is 37.5 ml. This is the sum of contractions from both gases: $C_1 + C_2 = 37.5$ ml $2V_1 + 1.5V_2 = 37.5$ (Equation 2)

4. Solving the System of Equations: We have two linear equations:

1. $V_1 + V_2 = 20$
2. $2V_1 + 1.5V_2 = 37.5$

From Equation 1, express $V_1$ in terms of $V_2$: $V_1 = 20 - V_2$

Substitute this expression for $V_1$ into Equation 2: $2(20 - V_2) + 1.5V_2 = 37.5$ $40 - 2V_2 + 1.5V_2 = 37.5$ $40 - 0.5V_2 = 37.5$ $0.5V_2 = 40 - 37.5$ $0.5V_2 = 2.5$ $V_2 = \frac{2.5}{0.5}$ $V_2 = 5$ ml

So, the volume of $C_2H_2$ in the mixture is 5 ml.

We can also find $V_1$: $V_1 = 20 - V_2 = 20 - 5 = 15$ ml.

Verification: Contraction from $C_2H_4 = 2V_1 = 2 \times 15 = 30$ ml. Contraction from $C_2H_2 = 1.5V_2 = 1.5 \times 5 = 7.5$ ml. Total contraction = $30 + 7.5 = 37.5$ ml, which matches the given information.

The final answer is $\boxed{5}$