Explain the chloride was treated with silver nitrate

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A metal, M, was converted to the chloride, MCl2. Then a solution of the chloride was treated with silver nitrate to give silver chloride crystals, which were filtered from the solution. MCl2(aq) + 2 AgNo3(aq) --> M(NO3)2(aq) + 2 AgCl(s) If 2.434 g of the metal 7.964 g of silver chloride, what is the atomic number of the metal? What is the metal?

Reference no: EM13170266

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