MCO3 ==> MO + CO2
0.88g CO2 = how many mols CO2? That will be 0.88/44 = 0.02 mols.
The equation tells you that mols MCO3 = 0.02 also.
Then mol = grams/molar mass You know g and you know mol, calculate molar mass. Then You know MCO3 molar mass. You know C and O. solve for atomic mass M. Finally identify M from the periodic table.
Suppose 2 grams of an unknown metal carbonate MCO3 is reacted with excess HCl and 0.88 grams of CO2 are produced. What is the atomic mass (molar mass) of the unknown metal M? What element is it?
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