10.0g of a compound, M2Cl2, solid reacts with NH3 in water to form equal amounts of elemental metal (4.25 g) and MNH2Cl

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answerhappygod
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10.0g of a compound, M2Cl2, solid reacts with NH3 in water to form equal amounts of elemental metal (4.25 g) and MNH2Cl

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10.0g of a compound, M2Cl2, solid reacts
with NH3 in water to form equal amounts of
elemental metal (4.25 g) and MNH2Cl (5.34 g) and
ammonium chloride (1.13 g). Calculate the relative atomic mass and
hence the identity of M.
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