When 2.54 g of solid iodine reacts with excess chlorine and the unreacted chlorine is evaporated, 4.67 g of a yellow product remains - VCE - SSCE Chemistry - Question 4 - 2007 - Paper 1

Next, the mass of the product can be determined by subtracting the mass of unreacted chlorine from the total mass:

$\text{Mass of Product} = \text{Mass of I} + \text{Mass of Cl} - \text{Mass of Chlorine remaining} = 2.54 \text{ g} + m_{Cl} - 4.67 \text{ g}$

Here, we must calculate the mass of chlorine that reacted.

The total mass before reaction is:

1. Mass of iodine (I) = 2.54 g
2. Moles of chlorine (Cl) reacting can be obtained from the mass of the remaining product, which is 4.67 g, so let’s denote this unknown mass of chlorine that reacted as ( m_{Cl} ).

From the empirical formula, we know that: $\text{Mass of I} + \text{Mass of Cl} = 4.67 \text{ g}$

Assuming the reacted mass of chlorine contributes to the yellow product, the empirical formula ratio can be determined after calculating the moles for iodine and chlorine. Using the molar mass of Cl (approximately 35.5 g/mol), we can find the moles of Cl.

1. Moles of Cl = $\frac{m_{Cl}}{35.5 \text{ g/mol}}$
2. The ratio ( n(I) : n(Cl) ) will give us the empirical formula.

After calculating moles and determining the ratio, we find that the empirical formula of the product is:

ICl₃ (Option B).