# More help on Stoich

1. Dec 14, 2007

### UWMpanther

[SOLVED] More help on Stoich

1. The problem statement, all variables and given/known data
Consider the reaction CaCO3 (s) -> CaO (s) + CO2 (g)
What mass of CaCO3 will produce 8.0L of CO2, measured at STP

I don't know where to begin.

2. Dec 14, 2007

### rocomath

what does STP mean? and what formula do you think you will need to use in order to relate, temperature, pressure, volume, and mass?

remember that

$$M=\frac{g}{mol}$$

so

$$n=\frac{g}{M}=\frac{g}{\frac{g}{mol}}$$

Last edited: Dec 14, 2007
3. Dec 14, 2007

### UWMpanther

stp meaning standard temp and pressure conditions.

4. Dec 14, 2007

### rocomath

Correct. And so we have 273.15K and 1.00atm.

From here, we need to evaluate the stoichiometric ratio between Calcium carbonate and Carbon dioxide.

5. Dec 14, 2007

### UWMpanther

Ok so the ratio would be

_g CaCO3 x 1mol/100gmol^-1 x 1mol CO2/1mol CaCO3 x 44gmol^-1/1 ?

6. Dec 14, 2007

### rocomath

our formula

$$PV=nRT$$

becomes

$$PV=\frac{mRT}{M}$$

solving for m (mass of Carbon dioxide)

$$m=\frac{MPV}{RT}$$

plugging in our known information ... STP, Volume of Carbon dioxide and it's Molar mass.

from there, you now know the mass of Carbon dioxide and compute how much Calcium carbonate is need to produce 8.0L of Carbon dioxide.

Last edited: Dec 14, 2007
7. Dec 14, 2007

### UWMpanther

Ok so its a multiple choice question. For the variable M would it be the CO2? And R is a constant if I'm not mistaken correct ( I believe .08206 L atm mol^-1)?

8. Dec 14, 2007

### rocomath

M = Molar mass of CO2

R = Gas constant (.08206 L*atm / K*mol)

9. Dec 14, 2007

### UWMpanther

I realized my error after I posted it. Thank you very much for the help!!!

Ok so then it is:

m= (100g mol CO2 x 1atm x 8L)/(273.15 x 0.08206 Latm/Kmol)

10. Dec 14, 2007

### rocomath

where did 100g mol CO2 come from?

Molar mass of CO2 is 44.01g/mol CO2

11. Dec 14, 2007

### UWMpanther

Ok here's how I solved for it:

m CO2 = (molar mass of CaCO3 x P x V)/(R x T)

after I entered the unkown's I get the answer 36g