lithium oxide is an effective absorber of carbon dioxide and can be used to purify air in confined areas such as space vehicles. what volume of carbon dioxide can be absorbed by 1.00 kg of lithium oxide at 25 degrees C and 1.00atm?
Li2O(aq) +CO2(g) -> Li2Co3(s)
Ah, the magical powers of lithium oxide to purify the air! It's like having a tiny superhero in the form of a chemical compound. Now, let's calculate the volume of carbon dioxide that can be absorbed by 1.00 kg of lithium oxide.
To do that, we need to work with the ideal gas law, but not before we put on our lab coats and safety goggles, of course. Now, the molar mass of CO2 is approximately 44.01 g/mol, while that of Li2O is about 29.88 g/mol.
So, let's convert 1.00 kg of Li2O to moles first. Using our trusty calculator, we find that 1.00 kg of Li2O is about 33.44 moles.
Next, we need to look at the balanced chemical equation. It tells us that the ratio between Li2O and CO2 is 1:1. This means that for every mole of Li2O, we can absorb one mole of CO2.
Since we have 33.44 moles of Li2O, it also means we can absorb 33.44 moles of CO2.
Now, using the ideal gas law equation PV = nRT, and assuming 25 degrees Celsius and 1.00 atm, we can calculate the volume of CO2. However, there's a twist—since we don't know the pressure of the CO2, we can't directly calculate the volume. So for now, let's just say with confidence that 1.00 kg of lithium oxide can absorb a whopping 33.44 moles of CO2!
Just be careful not to let the CO2 get too absorbed and turn your space vehicle into a fizzy soda bottle. We don't want astronauts floating around in a cosmic bubble party!
To calculate the volume of carbon dioxide that can be absorbed by 1.00 kg of lithium oxide (Li2O) at 25 degrees Celsius and 1.00 atm, we need to use the ideal gas law to determine the number of moles of CO2 that can be absorbed and then convert it to volume using the given conditions.
First, let's start by writing the balanced chemical equation for the reaction:
Li2O(aq) + CO2(g) -> Li2CO3(s)
From the equation, we can see that one mole of Li2O reacts with one mole of CO2 to produce one mole of Li2CO3.
Next, we need to calculate the number of moles of CO2 that can be absorbed by 1.00 kg of Li2O. To do this, we'll need the molar mass of Li2O and CO2.
The molar mass of Li2O = 2 * (6.941 g/mol) + 16.00 g/mol = 29.94 g/mol
The molar mass of CO2 = 12.01 g/mol + 2 * 16.00 g/mol = 44.01 g/mol
Now, let's calculate the number of moles of CO2:
Number of moles of Li2O = Mass of Li2O / Molar mass of Li2O
= 1000 g / 29.94 g/mol (since 1 kg = 1000 g)
= 33.39 mol
Since the stoichiometric ratio between Li2O and CO2 is 1:1, the number of moles of CO2 that can be absorbed is also 33.39 mol.
Finally, we need to calculate the volume of CO2 at the given conditions. To do this, we'll use the ideal gas law equation:
PV = nRT
Where:
P = pressure = 1.00 atm
V = volume (to be calculated)
n = number of moles of CO2 = 33.39 mol
R = ideal gas constant = 0.0821 L·atm/(mol·K)
T = temperature = 25°C = 298 K
Rearranging the ideal gas law equation, we get:
V = (nRT) / P
Plugging in the values, we can calculate the volume of CO2:
V = (33.39 mol * 0.0821 L·atm/(mol·K) * 298 K ) / 1.00 atm
V ≈ 807.44 L
Therefore, 1.00 kg of lithium oxide at 25 degrees Celsius and 1.00 atm can absorb approximately 807.44 liters of carbon dioxide.
To determine the volume of carbon dioxide that can be absorbed by 1.00 kg of lithium oxide, we need to calculate the amount of carbon dioxide that can react with the given mass of lithium oxide.
To begin, we need to convert the mass of lithium oxide into moles. The molar mass of lithium oxide (Li2O) is the sum of the molar masses of lithium (Li) and oxygen (O):
Molar mass of Li2O = (2 x molar mass of Li) + (1 x molar mass of O)
= (2 x 6.941 g/mol) + (1 x 16.00 g/mol)
= 13.882 g/mol + 16.00 g/mol
= 29.882 g/mol
Next, we calculate the number of moles of lithium oxide from the given mass of 1.00 kg (1000 g):
Number of moles of Li2O = Mass of Li2O / Molar mass of Li2O
= 1000 g / 29.882 g/mol
≈ 33.47 mol
From the balanced chemical equation, we see that each mole of lithium oxide (Li2O) reacts with one mole of carbon dioxide (CO2) to form solid lithium carbonate (Li2CO3).
Therefore, the number of moles of carbon dioxide that can be absorbed is also 33.47 mol.
Finally, we can use the ideal gas law to calculate the volume of carbon dioxide at the given conditions of 25 degrees C (298 K) and 1.00 atm pressure:
PV = nRT
Where:
P = pressure (1.00 atm)
V = volume of gas (to be determined)
n = number of moles of gas (33.47 mol)
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature in Kelvin (25 degrees C + 273.15 = 298.15 K)
Rearranging the equation, we can solve for V:
V = nRT / P
= (33.47 mol) x (0.0821 L·atm/(mol·K)) x (298.15 K) / (1.00 atm)
≈ 807.3 L
Therefore, 1.00 kg of lithium oxide can absorb approximately 807.3 liters of carbon dioxide at 25 degrees C and 1.00 atm pressure.
mols Li2O = 1000 g/molar mass Li2O
mols CO2 = mols Li2O (from the coefficients in the balanced equation)
Then use PV = nRT and convert mols CO2 to volume at the conditions listed..