what volume of carbon dioxide is produced when 49.7g of calcium carbonate reacts completely according to the following reaction at 25C and 1atm?
25C=298K
R= 8.31441 J K-1 mol-1
101325 x V= 49.7x8.31441x298
V= 49.7x8.31441x298
101325
=1.2153 m3
Please help am I even on the right track?
3 answers
I'm trying to find ____ liters carbon dioxide
whats the reaction?
CaCO3 => CaO + CO2 (Common Decomp Rxn) 1:1 rxn ratio
Ca. moles of CO2 from 49.7g CaCO3 then use Ideal Gas Law to Ca. Vol. CO2.
Moles CaCO3 = (49.7g/100g/mole)= 0.497 mole CaCO3 => 0.497 mole CO2.
From PV = nRT => V = nRT/P
n = 0.497 mole
R = 0.08206 L-Atm/mol-K
T = 298 K
P = 1 Atm
Substitute and solve. V = 12.2 L.
Note => when using the Ideal Gas Law, use R = 0.08206 L-Atm/mol-K. As such, data at the non-STP conditions must have units dimensions the same as that of the R-Value.
Ca. moles of CO2 from 49.7g CaCO3 then use Ideal Gas Law to Ca. Vol. CO2.
Moles CaCO3 = (49.7g/100g/mole)= 0.497 mole CaCO3 => 0.497 mole CO2.
From PV = nRT => V = nRT/P
n = 0.497 mole
R = 0.08206 L-Atm/mol-K
T = 298 K
P = 1 Atm
Substitute and solve. V = 12.2 L.
Note => when using the Ideal Gas Law, use R = 0.08206 L-Atm/mol-K. As such, data at the non-STP conditions must have units dimensions the same as that of the R-Value.
5 answers