Question

20 mL of a mixture of CO and $H_2$ were mixed with excess of $O_2$ and exploded & cooled. There was a volume contraction of 23 mL. All volume measurements corresponds to room temperature (27^\degree C) and one atmospheric pressure. Determine the volume ratio $V_1 : V_2$ of CO and $H_2$ in the original mixture

• A. 6.5 : 13.5
• B. 5 : 15
• C. 9:11
• D. 7:13

Answer 1

Answer: (D)

7:13

Answer 2

The problem involves the combustion of a gaseous mixture of CO and H2 with excess oxygen, followed by cooling. We need to determine the initial volume ratio of CO and H2 based on the total volume contraction.

Let $V_{CO}$ be the volume of CO and $V_{H_2}$ be the volume of H2 in the original mixture. Given: Total volume of the mixture = 20 mL $V_{CO} + V_{H_2} = 20$ mL --- (Equation 1)

The reactions involved are:

1. Combustion of CO: $CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)$ According to Gay-Lussac's Law of Gaseous Volumes (or Avogadro's Law), volumes are proportional to moles for gases at constant temperature and pressure. For $V_{CO}$ mL of CO: Volume of $O_2$ consumed = $\frac{1}{2} V_{CO}$ mL Volume of $CO_2$ formed = $V_{CO}$ mL Initial gaseous volume (reactants) = $V_{CO} + \frac{1}{2}V_{CO} = 1.5V_{CO}$ mL Final gaseous volume (product) = $V_{CO}$ mL Volume contraction due to CO reaction ($\Delta V_{CO}$) = Initial gaseous volume - Final gaseous volume $\Delta V_{CO} = 1.5V_{CO} - V_{CO} = 0.5V_{CO}$ mL

2. Combustion of H2: $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l)$ Since the mixture is cooled, water formed is in the liquid state, and its volume is negligible compared to gases. For $V_{H_2}$ mL of H2: Volume of $O_2$ consumed = $\frac{1}{2} V_{H_2}$ mL Volume of $H_2O$ formed = $V_{H_2}$ mL (liquid, so volume contribution to gas phase is 0) Initial gaseous volume (reactants) = $V_{H_2} + \frac{1}{2}V_{H_2} = 1.5V_{H_2}$ mL Final gaseous volume (product) = 0 mL Volume contraction due to H2 reaction ($\Delta V_{H_2}$) = Initial gaseous volume - Final gaseous volume $\Delta V_{H_2} = 1.5V_{H_2} - 0 = 1.5V_{H_2}$ mL

Given: Total volume contraction = 23 mL The total volume contraction is the sum of contractions from individual reactions: $\Delta V_{total} = \Delta V_{CO} + \Delta V_{H_2}$ $23 = 0.5V_{CO} + 1.5V_{H_2}$ --- (Equation 2)

Now we have a system of two linear equations:

1. $V_{CO} + V_{H_2} = 20$
2. $0.5V_{CO} + 1.5V_{H_2} = 23$

From Equation 1, express $V_{CO}$ in terms of $V_{H_2}$: $V_{CO} = 20 - V_{H_2}$

Substitute this into Equation 2: $0.5(20 - V_{H_2}) + 1.5V_{H_2} = 23$ $10 - 0.5V_{H_2} + 1.5V_{H_2} = 23$ $10 + V_{H_2} = 23$ $V_{H_2} = 23 - 10$ $V_{H_2} = 13$ mL

Now, substitute the value of $V_{H_2}$ back into Equation 1 to find $V_{CO}$: $V_{CO} = 20 - 13$ $V_{CO} = 7$ mL

The volume of CO is 7 mL and the volume of H2 is 13 mL. The volume ratio $V_{CO} : V_{H_2}$ is $7 : 13$.

The final answer is $\boxed{\text{7:13}}$.