One Gram of the oxide of a certain Metal is obtained by heating strongly 2.1g of the carbonate of the metal.
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One Gram of the oxide of a certain Metal is obtained by heating strongly 2.1g of the carbonate of the metal.

[From: ] [author: ] [Date: 11-05-12] [Hit: ]
?The difference in your problem corresponds to the loss of 1.1 g CO₂: (1.1 g)/(44 g/mol) = 0.and 0.025 mol of oxide remains in the metal oxide formed: (0.......
What is the equivalent Weight of the metal ??

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Write the conversion of carbonate to oxide and carbon dioxide:

CO₃²⁻ → CO₂ + O²⁻ and compare the corresponding masses: 60 g/mol → 44 g/mol + 16 g/mol

The difference in your problem corresponds to the loss of 1.1 g CO₂: (1.1 g)/(44 g/mol) = 0.025 mol
and 0.025 mol of oxide remains in the metal oxide formed: (0.025 mol O²⁻)(16 g/mol) = 0.4 g O²⁻

Therefore, there is 0.6 g of the metal in the metal oxide and the metal carbonate.

Since oxide has a 2– charge, its equivlaent weight is 8 g (the mass that takes up one mole of electrons when oxygen is reduced from ox. state 0 to 2–). Since we have 0.4 g of oxide, we have 0.05 equivalents and 0.05 equivlalents of the metal has a mass of 0.6 g. Hence the equivalent weight of the metal is (0.6 g)/((0.05 equivlaents) = 12 g/equivalent

On the face of it, there are two reasonable choices: magnesium or titanium, which would form MgO and TiO₂. (Approx. 12 g of Mg and Ti give one mole of electrons when they're respectively oxidized to Mg²⁺ and Ti⁴⁺). You can check that the information given works OK for:

MgCO₃ → CO₂ + MgO . . . or . . . Ti(CO₃)₂ → 2 CO₂ + TiO₂

In fact, I don't think there is any evidence that titanium(IV) carbonate actually exists while MgCO₃ is quite common.
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