2 AgNO3 + BaCl2 --> 2 AgCl + Ba(NO3)2 If you start with 25.00g of BaCl2 how many grams of AgCl will you end up with? (Assuming excess silver nitrate)

Answer 1

Assuming AgNO₃ is in excess, then the maximum amount of product (AgCl) will be determined by the amount of BaCl₂.

2AgNO₃(aq) + BaCl₂(aq) ==> 2AgCl(s) + Ba(NO₃)₂(aq) ... balanced equation

Using the stoichiometry of the balanced equation and dimensional analysis, we find the mass of AgCl:

25.00 g BaCl₂ x 1 mol BaCl₂/208.23 g x 2 mol AgCl/1 mol BaCl₂ x 143.32 g AgCl/mol = 34.41 g AgCl