Answer:
23.56
Explanation:
We know that BaCl₂ is the limiting reactant. We can write the following:
Grams BaCl₂ * Molar Mass of BaCl₂ * Mole ratio * Molar Mass of NaCl
41.97g BaCl₂ * (1 mol BaCl₂ / 208.23g BaCl₂) * (2 mol NaCl / 1 mol BaCl₂) * (58.44g NaCl / 1 mol NaCl) = 23.56g NaCl