# CO_2 gas, in the dry state, may be produced by heating calcium carbonate. CaCO_3 (s)DeltaCaO(s) + CO_2(g). What volume of CO_2, collected dry at 55 C and a pressure of 774 torr, is produced by complete thermal decomposition of 10.0 g of CaCO_3?

May 28, 2017

3.12 Liters if Dry Pressure is 656mmHg (Water Vapor Pressure - 118mmHg- is subtracted from 744mmHg), or 2.75 Liters if Dry Pressure is 744 mmHg. Calculation that follows is based upon 656mmHg dry pressure (=0.863Atm). Substitute 1.02Atm if 744mmHg is dry pressure.

#### Explanation:

10gramsCaCO_3(s) => CaO(s) + (?)O_2(g)

$\text{moles} C a C {O}_{3} = \left(\frac{10 g}{100 \left(\frac{g}{m o l}\right)}\right)$ = $0.10 \text{mole}$

From reaction ratios
=> moles $C a C {O}_{3} \left(s\right)$ consumed = moles ${O}_{2} \left(g\right)$ produced
=
0.10 mole $\left({O}_{2} \left(g\right)\right)$****

If a 'dry' volume is needed (VP_(H_2O)=118mmHg@55^oC), then the water vapor pressure at ${55}^{o} C$ is subtracted from the given 774mmHg pressure.
=> Dry Pressure = 774mmHg - 118mmHg = 656mmHg
656mmHg = (656mmHg)/(((760mmHg)/(Atm)) = 0.863Atm

Temperature = ${55}^{o} C + 273$ = 328K

R = (0.08206(((L)(Atm))/((mol)(K)))

Using the Ideal Gas Law => $P V = n R T \implies {V}_{\mathrm{dr} y} = \frac{n R T}{P}$

$V$ = $\frac{\left(0.10 m o l\right) \left(0.08206 \frac{\left(L\right) \left(A t m\right)}{\left(m o l\right) \left(K\right)}\right) \left(328 K\right)}{0.863 A t m} = 3.12 L$

OR => If dry pressure is 774torr = 1.02Atm => ${V}_{\mathrm{dr} y} = 2.75 L$