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