A mixture of two gases, 0.01 mol of C4H10(g) and 0.065 mol of O2(g), is pumped into a cylinder with — States of Matter Chemistry Question

STEPS:

1. Write and balance the chemical equation, identifying the states of matter:
The combustion of butane is represented by the balanced equation:
$$2\ \text{C}_4\text{H}_{10}(g) + 13\ \text{O}_2(g) \rightarrow 8\ \text{CO}_2(g) + 10\ \text{H}_2\text{O}(g)$$
It is crucial to note that every reactant and product in this reaction is a gas.

2. Check for a limiting reactant or stoichiometric mixture:
* The container is loaded with exactly $0.01\text{ mol of C}_4\text{H}_{10}(g)$ and $0.065\text{ mol of O}_2(g)$.
* The stoichiometric ratio of $\text{O}_2$ to $\text{C}_4\text{H}_{10}$ required by the balanced equation is $\frac{13}{2} = 6.5$.
* The ratio of reactants supplied in the experiment is $\frac{0.065\text{ mol}}{0.01\text{ mol}} = 6.5$.
* Because the supplied ratio perfectly matches the stoichiometric ratio, this is a stoichiometric mixture where both reactants are completely consumed without leaving any excess reactant behind.

3. Calculate the total moles of gas before and after the reaction:
* Initial moles ($n_{\text{initial}}$):
$$n_{\text{initial}} = 0.01\text{ mol of C}_4\text{H}_{10} + 0.065\text{ mol of O}_2 = \mathbf{0.075\text{ mol of reactant gas}}$$
* Final moles ($n_{\text{final}}$):
Using stoichiometric conversions from the balanced equation:
$$\text{Moles of CO}_2 = 0.01\text{ mol C}_4\text{H}_{10} \times \frac{8\text{ mol CO}_2}{2\text{ mol C}_4\text{H}_{10}} = 0.040\text{ mol}$$
$$\text{Moles of H}_2\text{O} = 0.01\text{ mol C}_4\text{H}_{10} \times \frac{10\text{ mol H}_2\text{O}}{2\text{ mol C}_4\text{H}_{10}} = 0.050\text{ mol}$$
$$n_{\text{final}} = 0.040\text{ mol of CO}_2 + 0.050\text{ mol of H}_2\text{O} = \mathbf{0.090\text{ mol of product gas}}$$
* Alternatively, we can simply observe that the balanced coefficients show that $15\text{ moles}$ of gas consumed ($2 + 13$) are replaced by $18\text{ moles}$ of gas produced ($8 + 10$). Thus, the reaction results in a net increase in the total number of gas molecules inside the cylinder ($n_{\text{final}} > n_{\text{initial}}$).

4. Relate the change in moles of gas to the cylinder volume (Ideal Gas Law):
* The gas constant ($R$) is constant, and the cylinder is returned to its original temperature ($20^\circ\text{C}$) and pressure ($1.0\text{ atm}$).
* According to the Ideal Gas Law ($PV = nRT$), volume is directly proportional to the number of moles of gas ($V \propto n$) when pressure and temperature are held constant.
* Because $n$ increased from $0.075\text{ mol}$ to $0.090\text{ mol}$, the volume of the gas mixture must expand to maintain constant pressure.

5. Determine the final position of the movable piston:
* To accommodate the larger volume of gas at $1.0\text{ atm}$, the gas must push the movable piston upward.
* Consequently, the piston will be higher than its original position because there are more moles of gas in the cylinder after the reaction, which is represented by Option A.

*

WHY_OTHERS_WRONG:

• Option B is incorrect: This option claims there are fewer moles of gas in the cylinder after the reaction, which violates stoichiometry. As calculated, $18\text{ moles}$ of gas are produced for every $15\text{ moles}$ consumed, so the total number of gas molecules increases, which must cause the piston to rise, not fall.
• Option C is incorrect: Although the final temperature and pressure are returned to their original values, the total number of gas molecules inside the cylinder has changed. Since there are more gas molecules under the same temperature and pressure, they must occupy a greater volume, forcing the piston to move upward.
• Option D is incorrect: Ideal gas behavior dictates that gas particles have negligible volume compared to the distance between them, meaning the relative physical sizes of $\text{CO}_2$ or $\text{H}_2\text{O}$ molecules compared to butane have no effect on the volume of the gas. What controls the volume is the absolute count of gas particles ($n$), which increases and forces the piston to rise.