Question

A vessel of volume $27 \times 10^{4}\text{ cc}$ contains a mixture of Hydrogen (molar mass $= 2\text{ g mol}^{-1}$) and Oxygen (molar mass $= 32\text{ g mol}^{-1}$) gas at standard temperature and pressure (S.T.P.). If the mass of hydrogen is $16\text{ g}$, find the mass of oxygen gas contained in the vessel.

• A. $64\text{ g}$
• B. $72\text{ g}$
• C. $129.7\text{ g}$
• D. $160\text{ g}$

Hint:

Always double-check that your volume is in liters or matching units before dividing by the S.T.P constant. Since $1\text{ L} = 1000\text{ cc}$, $27 \times 10^4\text{ cc} = 270\text{ L}$. Dividing $270$ by $22.4$ yields $\approx 12\text{ moles}$, which simplifies the calculation steps.

Answer 1

The Correct Option is C

Concept: According to Avogadro's law, $1\text{ mole}$ of any ideal gas fills an identical volume of $22.4\text{ Liters} = 22400\text{ cc}$ at standard temperature and pressure conditions. Dalton's Law of Partial Pressures states that the total number of moles in a container ($n_{\text{total}}$) is equal to the sum of the individual component moles: \[ n_{\text{total}} = n_{\text{hydrogen}} + n_{\text{oxygen}} \]

Step 1: Calculate the total moles from the vessel volume.
Given total volume $V = 27 \times 10^4\text{ cc} = 270000\text{ cc}$: \[ n_{\text{total}} = \frac{V}{22400} = \frac{270000}{22400} \approx 12.054\text{ moles} \]

Step 2: Find the number of moles of Hydrogen gas present.
Given mass $= 16\text{ g}$, molar mass $= 2\text{ g mol}^{-1}$: \[ n_{\text{hydrogen}} = \frac{\text{mass}}{\text{molar mass}} = \frac{16}{2} = 8\text{ moles} \]

Step 3: Deduce the mass of Oxygen gas present.
Using the total mole equation: \[ n_{\text{oxygen}} = n_{\text{total}} - n_{\text{hydrogen}} = 12.054 - 8 = 4.054\text{ moles} \] Converting moles of oxygen to grams (molar mass $= 32\text{ g mol}^{-1}$): \[ \text{Mass of oxygen} = 4.054 \times 32 \approx 129.73\text{ g} \]