Answer:
241.28 g NaCl(s)
Explanation:
Given 1.25 x 10²⁴ molecules of Cl₂ = 1.25 x 10²⁴molecules Cl₂(g)/6.023 x 10²³ molecules/mole = 2.08 mole Cl₂(g)
2Na(s) + Cl₂(g) => 2NaCl(s)
excess 2.08mole ? mole
∴moles NaCl(s) formed from 2.08 moles Cl₂(g) = 2(2.08 mole NaCl(s))
= 4.16 mole NaCl(g) = 4.16 mole NaCl(s) x 58 g NaCl(s)/mole NaCl(s) = 241.28 g NaCl(s)
