Direct reaction of solid iodine (I2) and gaseous chlorine (Cl2) produces an iodine chloride,(IxCly), a bright yellow solid. 0.678 g of iodine (I2) is mixed with 0.851 g of chlorine (Cl2). After the elements react, some chlorine is left over along with a single compound that contains the two

Question

elements (IxCly). The general unbalanced equation for this process is written below.

I2(s) + Cl2 (g) → IxCly (s) + (excess Cl2)

The excess chlorine is removed and allowed to react with elemental sodium producing 0.450 g of NaCl.
The balanced reaction is shown below.

2Na (s) + Cl2(g) → 2NaCl (s)

Based on this information, determine the empirical formula of the compound that was formed when
iodine reacted with chlorine.

I2 = 253.8 g/mol
Cl2 = 70.90 g/mol
NaCl = 58.44 g/mol

Answer 1

.........I2 + Cl2 ==> IxCly + xs Cl2
.....0.678g..0.851g....0..........x

2Na + Cl2 --> 2NaCl
You know NaCl weighed 0.450. Convert that to grams Cl2 and that is the excess amount. Subtract that from the 0.851 g there initially.
Now you have 0.678 g I2 and I obtained 0.578 g Cl2.

Now fine mols on the way to empirical formula
mols Cl2 = 0.578/35.44 = ?
mols I2 = 0.678/126.9 = ?
Now find the ratio with the smallest number being 1.00. The easy way to do that is to divide the smaller number by itself and divide the other number by the small number too. Then round to whole number.
Post your work if you get stuck. I ran through a quickie calculation and obtained ICl3

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