Question

A hydrocarbon containing C and H has 92.3% C. When 39 g of hydrocarbon was completely burnt in O_2, x moles of water and y moles of CO₂ were formed. x moles of water is sufficient to liberate 0.75 moles of H₂ with Na metal. What is the weight (in g) of oxygen consumed?

• A. 120
• B. 240
• C. 360
• D. 480

Hint:

For combustion reactions: - Balance carbon and hydrogen first. - Oxygen is adjusted based on reactant-product balance.

Answer 1

The Correct Option is A

Step 1: Determine the molecular formula of hydrocarbon 
- Since the hydrocarbon contains 92.3\% carbon, the remaining 7.7\% is hydrogen. 
- Let the molecular formula be \( C_xH_y \). 

Step 2: Find \( y \) from given data 
- The given condition states that the hydrocarbon produces \( x \) moles of water, which releases 0.75 moles of \( H_2 \) with sodium. 
- Since 1 mole of \( H_2O \) releases 1 mole of hydrogen atoms: 
\[ x = 2 \times 0.75 = 1.5 \] So, \( y = 3 \), leading to \( C_3H_3 \). 

Step 3: Calculate oxygen consumption - The balanced combustion reaction: 
\[ C_3H_3 + \frac{9}{4} O_2 \rightarrow 3 CO_2 + \frac{3}{2} H_2O \] - Molar mass of hydrocarbon = 39 g, and the oxygen required per mole is: 
\[ \frac{9}{4} \times 32 = 72 \text{ g} \]