[Duzzy]

A 14.6 g sample of the bicarbonate of the element X was heated and decomposed to
the oxide of bivalent metal, water and carbon dioxide. Indicate the formula of the
compound, assuming that the volume of CO2 (at STP) was 4.48 dm3.


How would I go about solving this problem? I know that the metal is going to be out of group IIa since it's bivalent but I'm stumped with the rest. Where would you start? Would you calculate the moles of CO2 by doing 4.48 (1/22.4) and then what? Once again I'm not looking for the answer because the answer won't really do me any good. I need to know how to approach these types of problems. Any help will be greatly appreciated.

[Borek]

Once you know number of moles of CO₂, can you calculate number of moles of the carbonate? Its molar mass?

[Duzzy]

So the mass of CO3 is 12g?

[Borek]

You can approach it this way as well.

[Duzzy]

Ya I'm lost. Can you give me another hint?

[Borek]

What is mass of the metal? How many moles of the metal?

[Duzzy]

If i knew the mass of the metal I wouldn't have this problem. I'm honestly lost. I haven't done this in forever. If it's 0.2 moles of CO3 is 12g/mol. with H2 it would be 12.4 and with one more O it would be 15.6 but the total weight is 14.6g. I'm missing something here and I have no idea what it is.

[Borek]

14.6 g sample minus 12 g of CO₃^2- - what is left?

[Duzzy]

2.6g

[Borek]

2.6 g of what?

1 a.m. here, I am going to bed.

[Duzzy]

Ok, so it comes out to MgO+h2o+ 2 co2


Correct me if I'm wrong

co2=4.48/2 since you get 2 CO2s out of every Mg(HCO3)2

so it's 0.1m for everything and CO2 is .2 m

so the metal x is Mg since 2.4/.1=24

[Borek]

It is not entirely clear to me what you did, could be you made two errors that canceled out - but yes, Mg is the correct answer.