How many grams of barium chloride are produced when 1.25x10^23 molecules of HCl reacts with barium according to the following equation?

Answer 1

You didn't show "the following equation". So, I'll assume it is one of the following:

Ba + 2HCl ==> BaCl₂ + H₂

Ba(OH)₂ + 2HCl ==> BaCl₂ + 2H₂O

Either equation will provide the same answer, so I'll use the first one.

moles HCl present = 1.25x10²³ molecules HCl x 1 mol HCl / 6.022x10²³ molecules = 0.208 moles HCl

moles BaCl₂ formed = 0.208 mols HCl x 1 mol BaCl₂ / 2 moles HCl = 0.104 mols BaCl₂ formed

mass BaCl₂ formed = 0.104 mols BaCl₂ x 208 g BaCl₂ / mol = 21.6 g BaCl₂ formed