[Miss1nik2]

A 20.29 g sample of impure MgCO3 is completely decomposed at 1000oC in previously evacuated 2.0 L reaction vessel. After the reaction was complete the solid residue had a mass of 15.9 g. Assume that only MgCO3 could produce gas CO2 what percent of original sample was magnesium carbonate? What was the pressure of the CO2 produced?

So far I have found that 20.29g MgCO3 makes 10.59g CO2 or .24 mol.

And I know that the equation that I need to use is PV/T = PV/T. But I am stuck after that.

[Mitch]

See #1 n signature

[Miss1nik2]

I got further in the problem...

(a) 15.9/20.29= 78.54%

and for part (b)

I found that 20.29g of MgCO3 makes .24mol CO2. So....

Pressure of CO2 = 12.54 atm.

Would anyone be willing to check my work?? THANK YOU!

[Borek]

(a) 15.9/20.29= 78.54%

That will work if 15.9 were the mass of magnesium carbonate. This is mass of residue, magnesium carbonate decomposed, so it can't be present in the residue.

Hint: where is the missing mass?

[Miss1nik2]

I have actually been talking to a tutor who said that exact thing about this problem. My teacher (unfortunately) very often gives problems with parts missing. The problem that I posted is the entire problem.

I was also wondering if I did part (b) right.

I used 20.29g of MgCO3 to find .24mol CO2 and that the Pressure of CO2 = 12.54 atm.

But was that wrong? Should I have used 15.9g of MgO to find .39 mol, which would then give you P = 20.77 atm.

Thank you again!

[sjb]

I used 20.29g of MgCO3 to find .24mol CO2 and that the Pressure of CO2 = 12.54 atm.

But was that wrong? Should I have used 15.9g of MgO to find .39 mol, which would then give you P = 20.77 atm.

In my opinion, neither.

You are told that the sample of MgCO₃ is impure, so you cannot tell how many moles of that you have, and so how many moles of CO₂ "should" be produced.

Similarly, the MgO is also impure, and so also cannot give you a mass of CO₂. So what does that leave? (I think you will need to assume that CO₂ is the only gas given off....)