Final answer:
Before and after the reaction between solid sodium and chlorine gas, there are four atoms present. Two sodium atoms combine with two chlorine atoms from one chlorine molecule, creating two molecules of sodium chloride, and conserving the total number of atoms.
Step-by-step explanation:
During the combination of solid sodium and chlorine gas, two sodium atoms interact with one molecule of chlorine gas (which consists of two chlorine atoms), resulting in two molecules of sodium chloride. Prior to the reaction, there are four atoms in total: 2 sodium atoms (from solid sodium) and 2 chlorine atoms (from a single chlorine gas molecule, Cl₂).
After the reaction, these atoms rearrange to form two molecules of sodium chloride (NaCl), each molecule consisting of one sodium atom and one chlorine atom, totaling to four atoms post-reaction.
The balanced chemical equation for this reaction is:
2Na(s) + Cl₂(g) → 2NaCl(s)
This indicates that the number of atoms remains the same before and after the reaction, maintaining the law of conservation of mass. Each sodium atom donates one electron to each chlorine atom, forming sodium cations (Na⁺) and chloride anions (Cl⁻), which then bond to form the ionic compound NaCl. The initial and final atom count is therefore the same.