Question

One molecule of lodine (Is) and two molecules of sodium chloride (NaCI) combine in a reaction. How many atoms are in the products?

Answer 1

To determine the total number of atoms in the products of the reaction between one molecule of iodine (I2) and two molecules of sodium chloride (NaCl), we first need to look at the reactants.

1. **Iodine (I2)**: This molecule contains 2 iodine atoms.
2. **Sodium Chloride (NaCl)**: Each molecule of NaCl contains 1 sodium atom and 1 chlorine atom. Since there are 2 molecules of NaCl, we have:
- 2 sodium atoms (1 from each NaCl)
- 2 chlorine atoms (1 from each NaCl)

Now let's add up the total number of atoms from all the reactants:

- From iodine (I2): 2 iodine atoms
- From 2 NaCl: 2 sodium atoms + 2 chlorine atoms = 4 atoms

Now, combining these results:
- Total atoms = 2 (from I2) + 4 (from 2 NaCl) = 6 atoms

Assuming a reaction where iodine combines with sodium chloride, the products can typically be sodium iodide (NaI) in a 1:1 stoichiometric ratio with the sodium and iodine. However, for calculating the total number of atoms, regardless of the specific products formed, the total count remains 6.

Thus, the total number of atoms in the products is **6 atoms**.