To solve this problem, we first need to determine the number of moles of CO2 present in 50L at 25 degrees Celsius and 1.0 atm:
PV = nRT
(1.0 atm)(50 L) = n(0.0821 L.atm/mol.K)(298K)
n = 2 moles
From the balanced chemical equation, we see that it takes 4 moles of KO2 to react with 2 moles of CO2:
4 moles KO2 : 2 moles CO2 = x moles KO2 : 2 moles CO2
x = 4 moles KO2
Now we need to calculate the molar mass of KO2:
The molar mass of potassium is 39.10 g/mol, oxygen is 16.00 g/mol, and there are 2 oxygen atoms in KO2, so the molar mass of KO2 is:
39.10 + (16.00 x 2) = 71.10 g/mol
Now, we can calculate the mass of KO2 needed to react with 2 moles of CO2:
4 moles x 71.10 g/mol = 284.40 g
Finally, convert this mass to kilograms by dividing by 1000:
284.40 g / 1000 = 0.2844 kg
Therefore, the mass of KO2 needed to react with 50L of CO2 at 25 degrees Celsius and 1.0 atm is 0.2844 kg.
Regarding the reaction below
4KO2 (s) + 2CO2 (g) = 2K2CO3(s) + 3O2(g)
Calculate the mass of KO2 needed to react with 50L of CO2 at 25 degrees Celsius and 1.0 atm.convert your final answer to kilo grams.
1 answer