Final answer:
In a reaction to produce sodium chloride from 8 g of sodium and 8 g of diatomic chlorine, chlorine (Cl₂) is the limiting reactant because it will be consumed first. The correct option is b).
Step-by-step explanation:
The limiting reactant in a chemical reaction is the reactant that is entirely consumed first and thus determines the maximum amount of product that can be formed. To find the limiting reactant between sodium (Na) and diatomic chlorine (Cl₂) in the production of sodium chloride, we first write the balanced chemical equation:
2 Na(s) + Cl₂(g) → 2 NaCl(s)
The balanced equation indicates that two moles of sodium react with one mole of diatomic chlorine to produce two moles of sodium chloride. Next, we convert the given masses to moles using the molar masses of the reactants (Na: approx. 23 g/mol, Cl₂: approx. 71 g/mol) to see which reactant is in a smaller stoichiometric amount:
- For sodium: 8 g ÷ 23 g/mol = 0.348 mol Na
- For diatomic chlorine: 8 g ÷ 71 g/mol = 0.113 mol Cl₂
Comparing these amounts to the stoichiometry of the reaction, we see that 0.348 mol Na would require 0.174 mol Cl₂ to react completely, but we only have 0.113 mol Cl₂ available. Therefore, chlorine (Cl₂) is the limiting reactant because it will be consumed first, and some amount of sodium will remain unreacted. Option b) is the correct one.