Question

A hydrocarbon has 85.7% of carbon by weight. 56 g (2 moles) of hydrocarbon was completely burnt in oxygen and obtained \( CO_2 \) and \( H_2O \). What is the weight (in g) of \( CO_2 \) formed? - (C = 12 u, H = 1 u, O = 16 u)

• A. \(176\)
• B. \(88\)
• C. \(132\)
• D. \(352\)

Hint:

To determine the mass of \( CO_2 \) produced, use the ratio: \[ \text{Mass of } CO_2 = \frac{\text{Molar mass of } CO_2}{\text{Atomic mass of C}} \times \text{Mass of Carbon in hydrocarbon} \]

Answer 1

The Correct Option is A

Step 1: Determining the Carbon Mass
- The hydrocarbon contains 85.7\% carbon by weight.
- Given hydrocarbon mass = 56 g
- Mass of carbon in the hydrocarbon: \[ \text{Mass of Carbon} = \frac{85.7}{100} \times 56 = 48 \text{ g} \] Step 2: Mass of \( CO_2 \) Formed
- Each mole of carbon (\( C \)) burns completely to form 1 mole of \( CO_2 \).
- Mass of 1 mole of carbon = 12 g
- Mass of 1 mole of \( CO_2 \) = 44 g Using proportion: \[ \text{If 12 g of Carbon produces 44 g of } CO_2 \] \[ \text{Then, 48 g of Carbon produces} \quad \frac{44}{12} \times 48 \] \[ = 176 \text{ g of } CO_2 \] Thus, the correct answer is \( \mathbf{(1)} \ 176 \).