Excess silver(I) nitrate was added to a 8.500 g mixture containing some amount of barium chloride, and 7.123 g of silver chloride was obtained. The unbalanced equation is AgNO3 + BaCl2(aq) --> Ba(NO3)2(aq) + AgCl(s).
What is the mass% of BaCl2 in the mixture?

60.88%

Explanation:

The balanced equation of the reaction is given as;

2 AgNO3 (aq) + BaCl2 (aq) → 2 AgCl (s) + Ba(NO3)2 (aq)

Since AgNO3 is in excess, the limiting reactant is BaCl2. From the reaction;

1 mol of BaCl2 produces 2 mol of AgCl

Converting to masses;

Mass = Number of mol * Molar mass

BaCl2;

Mass = 1 * 208.23 g/mol = 208.23 g

AgCl;

Mass = 2 * 143.32 g/mol = 286.64 g

208.23 g BaCl2 produces 286.64 g of AgCl

x g BaCl2 produces 7.123 g of AgCl

Solving for x;

x = 7.123 * 208.23 / 286.64 = 5.1745 g

Mass percent = Mass / Total mass of Mixture  * 100

Mass Percent = 5.1745 / 8.500 = 0.6088 * 100 = 60.88%

i have no clue hope you find the answer

the right answer is c i just did the question.