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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
Question 4
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. The empirical formula of the pr... show full transcript
Worked Solution & Example Answer: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
Step 1
Calculate Moles of Iodine
Answer
To begin, we need to calculate the number of moles of solid iodine (I₂) that reacts. The molar mass of iodine (I) is approximately 126.9 g/mol. Since I₂ has a molar mass of The number of moles of iodine can be calculated using the formula: n = rac{mass}{molar ext{ mass}} Substituting the values: n = rac{2.54 ext{ g}}{253.8 ext{ g/mol}} \\ n ext{ of I₂} ≈ 0.0100 ext{ mol}
Step 2
Calculate Moles of Yellow Product
Answer
Next, we determine the number of moles in the yellow product that remains after the reaction. Given that 4.67 g of the product is left: First, we need the molar mass of the yellow product. Assuming the product is ICl, Now, we calculate moles of the product: n = rac{4.67 ext{ g}}{162.4 ext{ g/mol}} ≈ 0.0287 ext{ mol}
Step 3
Determine Empirical Formula
Answer
Now, we can find the ratio of moles of iodine to moles of chlorine. For the yellow product (ICl), it consists of 1 mole of I for every mole of Cl. Therefore, the empirical formula can be derived:
- Moles of I = 0.0100 mol
- Moles of Cl in ICl = 0.0287 mol
To find the simplest ratio, we divide both by the smallest number of moles, which is 0.0100:
- I: rac{0.0100}{0.0100} = 1
- Cl: rac{0.0287}{0.0100} ≈ 2.87.
Approximating Cl to the nearest integer, we see it is closest to 3. Thus, the empirical formula is ICl₃ (option B).