Question

Can you help me figure out if i am doing this correctly? I don't want the answer, just to know if i'm on the right track.

Answer 1

Group 2A Metal carbonates are decomposed to the metal oxide and Co2 on heating. When 0.158g of group 2A metal carbonate was heated, evolved CO2 has a pressure of 69.8 mmHg in a 285 mL flask at 25C. Determine the formula weight of the metal carbonate.

a) Write a balanced equation after determining the metal.
b)using heats of formation from appendix, and heat of formation of Metal oxide (=-592kJ/mol), determine the amount of head absorbed/released for 0.158g of the metal carbonate?


so i wrote this equation first
\[MCO_3 \rightarrow MO + CO_2 \]
0.158g MCO3 69.8mmHg = .0918 atm
285mL = .285 L
25C = 298 K

i plugged those into PV=nRT (\[n=\frac{ PV }{ RT }\])
\[n= \frac{ (.0918)(.285) }{ (.0821)(298) }\] and n= .00107...

since it is 1:1 ratio MCO3 is also .00107mol so i did g/mol to get the MM which is (.158g/.00107) = 147.66g/mol

Not sure this is correct though, because the element closest to this MM is plutonium. But the closest in Group 2A is Barium.

Answer 2

Keeping up with using Barium
for

a) i got BaCo3 ---> BaO + CO2

Answer 3

@Owlcoffee are you able to help with this.

Answer 4

Makes sense to me, I think you are on the right track here.

Answer 5

Ok, thank you! I'm kind of stuck on b) though. i know i do the delta H for the balanced equation in a). But it also says that the Hformation of metal oxide is -592 kJ/mol

Answer 6

But in the appendix BaO(s) is -558.2 kJ/mol, CO2 is -393.5, and BaCO3 is -1218.8

Answer 7

and the Delta H for that reaction comes out to be 267.1 kJ/mol

Answer 8

Hmmm... Not so sure about what to say about that one, did you make sure the calculations are correct?

Answer 9

Yes, i've done them a few times.

Answer 10

Well, it should be okay, now I became confused as well, the difference isn't so big, is it?

Answer 11

Well the question says the the Delta H (formation) of Metal Oxide is -592 kJ/mole and that is the part that is really confusing me.

Answer 12

Ok, so if i use the heat of formation of metal oxide number as in place of -558.2 (for BaO; Barium Oxide) i get a Delta H of 233.3

Answer 13

I figured out what i did incorrectly! It is not Ba, it is Sr! i forgot to -60g for the CO3... so when i get to delta H of rxn it is 232.88kJ/mol, now do i just *that by the moles found to find the heat absorbed/released for the .158g?

Answer 14

you're on the right track, keep it up

Answer 15

Awesome! so is the Delta H value i'm looking for .24918? @jfraser

Answer 16

without doing out all the math again, i'd say you're correct

Answer 17

Awesome! Thanks :)