Question

A hydrocarbon containing C and H has 92.3\% of C. When 52 g of hydrocarbon is completely burnt in oxygen, \( x \) moles of water and \( y \) moles of CO\(_2\) were formed. The liberated water is sufficient to liberate one mole of H\(_2\) when reacted with sodium metal. What is the weight (in g) of O\(_2\) consumed?

• A. \( 80 \) g
• B. \( 160 \) g
• C. \( 240 \) g
• D. \( 320 \) g

Hint:

- To find the oxygen required, first determine the amount of CO\(_2\) and H\(_2\)O produced from combustion. - Use the reaction equations to relate the moles of oxygen consumed. - Always check units and molecular masses while solving combustion problems.

Answer 1

The Correct Option is B

Step 1: Determine the mass of Carbon and Hydrogen in the hydrocarbon Given that the hydrocarbon contains 92.3\% Carbon, the mass of Carbon in 52 g of the hydrocarbon is: \[ \text{Mass of Carbon} = \frac{92.3}{100} \times 52 = 48 \text{ g} \] Since the hydrocarbon consists of only Carbon and Hydrogen, the remaining mass is Hydrogen: \[ \text{Mass of Hydrogen} = 52 - 48 = 4 \text{ g} \] Step 2: Determine the number of moles of CO\(_2\) and H\(_2\)O formed The number of moles of CO\(_2\) formed from the complete combustion of Carbon: \[ \text{Moles of CO}_2 = \frac{\text{Mass of Carbon}}{\text{Molar mass of Carbon}} = \frac{48}{12} = 4 \text{ moles} \] The number of moles of H\(_2\)O formed from the complete combustion of Hydrogen: \[ \text{Moles of H}_2O = \frac{\text{Mass of Hydrogen}}{\text{Molar mass of Hydrogen in H}_2O} = \frac{4}{2} = 2 \text{ moles} \] Step 3: Calculate the mass of O\(_2\) consumed Using the reaction equation: \[ \text{C} + O_2 \rightarrow CO_2 \] \[ \text{H}_2 + \frac{1}{2} O_2 \rightarrow H_2O \] For the 4 moles of CO\(_2\) produced, the required oxygen is: \[ \text{Moles of O}_2 = 4 \] For the 2 moles of H\(_2\)O produced, the required oxygen is: \[ \text{Moles of O}_2 = 1 \] Total oxygen moles: \[ \text{Total Moles of O}_2 = 4 + 1 = 5 \] Since 1 mole of O\(_2\) weighs 32 g, the total mass of oxygen required is: \[ \text{Mass of O}_2 = 5 \times 32 = 160 \text{ g} \]