Chemical calculations Q

[GuppyR]

Could anyone please help me with the reaction between BaO

[exchemist]

  On 7/2/2022 at 6:25 PM, GuppyR said:

Could anyone please help me with the reaction between BaO

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How many moles of CO2 are driven off?

[GuppyR]

  On 7/3/2022 at 8:47 AM, exchemist said:

How many moles of CO2 are driven off?

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Oh so I can find the number of moles of CO2 by calculating the mass difference and apply that to the stoichiometry of the XO formed right.

Afterwards so could I find the molar mass of the whole compound XO formed by {mass of residue (XO)} X {no. Of moles} of it?

 

[exchemist]

  On 7/3/2022 at 2:32 PM, GuppyR said:

Oh so I can find the number of moles of CO2 by calculating the mass difference and apply that to the stoichiometry of the XO formed right.

Afterwards so could I find the molar mass of the whole compound XO formed by {mass of residue (XO)} X {no. Of moles} of it?

 

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You will find the number of moles is a suspiciously convenient number....... 

But your residue is a mixture of oxides: BaO and XO. (You are told it involves XCO3, not X2CO3 or X2(CO3)3 or something, so X has to be divalent and presumably produces an oxide with formula XO.) You now know how many moles of X you have but not how many of Ba, yet.

That's where the acid reaction comes in, I think. I have not worked it through myself but, looking at the problem, I think that's what you are supposed to use to find the answer.

Good luck.        

[John Ryan]

  On 7/3/2022 at 2:48 PM, exchemist said:

You will find the number of moles is a suspiciously convenient number....... 

But your residue is a mixture of oxides: BaO and XO. (You are told it involves XCO3, not X2CO3 or X2(CO3)3 or something, so X has to be divalent and presumably produces an oxide with formula XO.) You now know how many moles of X you have but not how many of Ba, yet.

That's where the acid reaction comes in, I think. I have not worked it through myself but, looking at the problem, I think that's what you are supposed to use to find the answer.

Good luck.        

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Thanks for the advice. I had a similar chemistry homework, it helped me figure it out.