$CO_2$ 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 $CaCO_3$?

Md Ariful Islam

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Md Ariful Islam

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

$"moles" CaCO_3 = ((10g)/(100(g/(mol))))$ = $0.10"mole"$

From reaction ratios
=> moles $CaCO_3(s)$ consumed = moles $O_2(g)$ produced
= 0.10 mole $(O_2(g))$****

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

Temperature = $55^oC + 273$ = 328K

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

Using the => $PV = nRT => V_(dry) = (nRT)/(P)$

$V$ = $((0.10mol)(0.08206((L)(Atm))/((mol)(K)))(328K))/(0.863Atm) = 3.12L$

OR => If dry pressure is 774torr = 1.02Atm => $V_(dry) = 2.75L$

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