To solve this problem, we can use the Ideal Gas Law to convert the volume of carbon dioxide to moles, then use the stoichiometry of the reaction to determine the amount of KO2 required.
First, we need to calculate the number of moles of CO2 present in 50 L at 25°C and 1.0 atm:
PV = nRT
(1.0 atm)(50 L) = n(0.0821 L.atm/mol.K)(298 K)
n = 2.04 moles of CO2
Next, we use the stoichiometry of the reaction to determine the amount of KO2 needed. From the balanced equation, we see that 4 moles of KO2 react with 2 moles of CO2:
4 moles KO2 reacts with 2 moles CO2
Therefore, 2 moles KO2 reacts with 1 mole CO2
Thus, the number of moles of KO2 needed is 2.04/2 = 1.02 moles
Finally, we calculate the mass of KO2 needed:
Molar mass of KO2 = 71.12 g/mol
Mass = moles * molar mass
Mass = 1.02 mol * 71.12 g/mol = 73.02 g
Convert this mass to kilograms:
73.02 g = 0.07302 kg
Therefore, the mass of KO2 needed to react with 50 L of carbon dioxide at 25oC and 1.0 atm is 0.07302 kg.
The carbon dioxide generated by the personnel in the artificial atmosphere of submarines and
spacecraft must be removed from the air and the oxygen recovered. Submarine design teams
have investigated the use of potassium superoxide (KO2) as an air purifier because this
compound reacts with carbon dioxide and releases oxygen
4KO2(s) + 2CO2(g) 2K2CO3(s) + 3O2(g).
Calculate the mass of KO2 needed to react with 50 L of carbon dioxide at 25oC and 1.0 atm.
Convert your final answer to kilograms.
1 answer