Pd(NO₃)₂ + NaCl ---> PdCl₂ + NaNO₃ 140 grams of lead (II) nitrate (Pd(NO₃)₂) are dissolved in water then mixed with a solution containing 75 grams of sodium chloride (NaCl). The reaction creates the precipitate, lead (II) Chloride (PdCl₂), and a solution of sodium nitrate (NaNO₃) . How many grams of lead (II) Chloride can be produced given the initial amount of lead (II) nitrate?

Stoichometry based.

All I need is the set up, but you can give answer if you want.

Please help.