To determine the number of atoms in the products of the reaction between one molecule of iodine (I\(_2\)) and two molecules of sodium chloride (NaCl), let's break it down:
-
Iodine (I\(_2\)): Each molecule of I\(_2\) contains 2 iodine atoms. So, for one molecule of I\(_2\):
- Iodine atoms = 2
-
Sodium Chloride (NaCl): Each molecule of NaCl contains 2 atoms (1 sodium atom and 1 chlorine atom). So for two molecules of NaCl:
- NaCl molecules = 2
- Sodium atoms from 2 NaCl = 2
- Chlorine atoms from 2 NaCl = 2
So, the total number of atoms in the products is:
- From I\(_2\): 2 iodine atoms
- From 2 NaCl: 2 sodium atoms + 2 chlorine atoms = 4 atoms
Adding these up gives: 2 (from I\(_2\)) + 2 (sodium from NaCl) + 2 (chlorine from NaCl) = 6 atoms total.
Therefore, the total number of atoms in the products is 6.
The correct answer is b. 6.