Answer to Question #134119 in Organic Chemistry for Inva

Question

We pour a 10l mixture of He, CO2 and CH4 in a container with NaOH. The volume of the gas that comes out is 7,24l. This gas that comes out reacts with 0.4 mol of Cl in presence of UV rays to give CCl4. Write the chemical reactions and determine the percentage of each ingredient in the 10l mixture.

Answer 1

Out of these three gases, only CO₂ reacts with NaOH:

2NaOH + CO₂ --> Na₂CO₃ + H₂O

Therefore, the volume and percentage of CO₂ can be determined directly:

V(CO₂) = 10L - 7.24L = 2.76 L

%CO₂ = (2.76L / 10L) * 100% = 27.6%

The gas that reacts with Cl₂ (there must be a mistake in the task description as chlorine exists only as a diatomic molecule) is CH₄.

CH₄ + 4Cl₂ --> CCl₄ + 4HCl

According to the reaction, 4 moles of Cl₂ react with 1 mole of CH₄. Therefore, 0.4 moles of Cl₂ would react with 0.1 mole of CH₄. Now the volume and percentage of CH₄ can be determined (assuming STP conditions):

V (CH₄) = n * V_M = 0.1mol * 22.4L/mol = 2.24 L

%CH₄ = (2.24L / 10L) * 100% = 22.4%

Finally, the remaining gas is He:

%He = 100% - 27.6% - 22.4% = 50%

Answer:

%CO₂ = 27.6%

%CH₄ = 22.4%

%He = 50%