Question

The molar composition of a gas is $10\%\text{ }\text{H}_2$, $10\%\text{ }\text{O}_2$, $30\%\text{ }\text{CO}_2$ and balance $\text{H}_2\text{O}$. If $50\%$ of the $\text{H}_2\text{O}$ condenses, the final mole percent of $\text{H}_2$ in the gas on a dry basis will be:

• A. $10\%$
• B. $5\%$
• C. $18.18\%$
• D. $20\%$

Hint:

Be careful of distractor details! The mention of "$50\%$ condensation" is extra information meant to test your conceptual clarity. A "dry basis" calculation ignores water content entirely, making the condensation step irrelevant.

Answer 1

The Correct Option is D

Concept: The terms Dry Basis and Wet Basis specify how component percentages are expressed relative to the presence of water vapor:
• Wet Basis: Includes all components present in the gas mixture, including water vapor ($\text{H}_2\text{O}$).
• Dry Basis: Calculates component percentages by completely omitting the water vapor content from the denominator. Therefore, changes in the amount of water due to condensation or evaporation do not affect calculations performed on a strict dry basis.

Step 1: Analyze the original composition and select a basis.
Let us assume a baseline of $100\text{ moles}$ of the original wet gas mixture. The initial layout of components is: \[ n_{\text{H}_2} = 10\text{ moles} \] \[ n_{\text{O}_2} = 10\text{ moles} \] \[ n_{\text{CO}_2} = 30\text{ moles} \] The remaining balance is $\text{H}_2\text{O}$. Thus, the initial number of water moles is: \[ n_{\text{H}_2\text{O}} = 100 - (10 + 10 + 30) = 100 - 50 = 50\text{ moles} \]

Step 2: Understand the implication of the "Dry Basis" requirement.
The problem asks for the mole percent of $\text{H}_2$ on a dry basis. By definition, calculating a composition on a dry basis means we ignore the water content entirely, regardless of how much $\text{H}_2\text{O}$ condenses. Therefore, the total number of non-water moles is: \[ n_{\text{dry}} = n_{\text{H}_2} + n_{\text{O}_2} + n_{\text{CO}_2} \] \[ n_{\text{dry}} = 10 + 10 + 30 = 50\text{ moles} \]

Step 3: Calculate the dry-basis mole percent of $\text{H}_2$.
Using the dry components layout: \[ \text{Mole \% of }\text{H}_2\text{ (Dry Basis)} = \left( \frac{n_{\text{H}_2}}{n_{\text{dry}}} \right) \times 100 \] \[ = \left( \frac{10}{50} \right) \times 100 = 0.2 \times 100 = 20\% \]