You mix 55 ml of 1.00 m silver nitrate with 25 ml of 1.06 m sodium chloride. what mass of silver chloride should you form?

AgNO3 + NaCl ---> AgCl(s) + NaNO3

find moles of each

mmol = M*V

mmol = 55*1 = 55 mmol of Ag+

mmol = 25 * 1.06 = 26.5 mmol of Cl-

then limitn greactant is Cl

we can produce up to 26.5 mmol of Agcl

mass = mol*MW = (26.5 * 10^-3) *  143.32 = 3.7980 g of AgCl

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