What is the volume of the oxygen produced when 1.00L carbon dioxide reacts with 20.0 g of Na2O2 based on the following unbalanced reaction at 25C and 2.00 bar?
Na2O2 + CO2 -> O2 + Na2CO3
R = 0.083145 L bar K-1 mol-1
I got 1.57L but the answer is 500mL
It would have been so much easier if you had shown your work; then I could have found your error.
First, you didn't balanced the equation.
2Na2O2 + 2CO2 ==> O2 + 2Na2CO3
mols Na2O2 = grams/molar mass = ? I have approx 0.3 but that's just an estimate.
Use PV = nRT and solve for mols CO2. I have approx 0.04, again an estimates as are all of the other numbers I quote.
Did you calculate that CO2 was the limiting reagent. It is.
Using the coefficients in the balanced equation, convert mols CO2 to mols O2. I get approx 0.02
Then use PV = nRT and calculate volume. I get 0.5L which is 500 cc.
To find the volume of oxygen produced when 1.00L of carbon dioxide reacts with 20.0g of Na2O2, you need to use the stoichiometry of the balanced chemical equation.
1. Start by balancing the given chemical equation:
Na2O2 + CO2 -> O2 + Na2CO3
After balancing the equation, it becomes:
2 Na2O2 + 2 CO2 -> O2 + 2 Na2CO3
2. Calculate the number of moles of Na2O2 and CO2:
Molecular mass of Na2O2 = (2 x atomic mass of Na) + (2 x atomic mass of O) = (2 x 22.99 g/mol) + (2 x 16.00 g/mol) = 77.98 g/mol
Number of moles of Na2O2 = mass / molar mass = 20.0 g / 77.98 g/mol = 0.256 mol
Molecular mass of CO2 = (atomic mass of C) + (2 x atomic mass of O) = 12.01 g/mol + (2 x 16.00 g/mol) = 44.01 g/mol
Number of moles of CO2 = volume / molar volume = 1.00 L / 22.41 L/mol (at STP) * (273 K / 298 K) * (2.00 bar / 1.00 bar) = 0.546 mol
3. Use the stoichiometry of the balanced equation to determine the number of moles of oxygen produced:
From the balanced equation, we can see that 2 moles of Na2O2 react to produce 1 mole of O2. Therefore, the number of moles of O2 produced is half the number of moles of Na2O2 used.
Number of moles of O2 produced = 0.256 mol x (1/2) = 0.128 mol
4. Calculate the volume of O2 produced using the ideal gas law:
PV = nRT
Where:
P = pressure (2.00 bar)
V = volume
n = number of moles (0.128 mol)
R = ideal gas constant (0.083145 L bar K-1 mol-1)
T = temperature (25°C = 298 K)
Rearranging the equation, we get:
V = (nRT) / P
V = (0.128 mol x 0.083145 L bar K-1 mol-1 x 298 K) / 2.00 bar
V = 0.3178 L
5. Convert the volume from liters to milliliters:
0.3178 L x 1000 ml/L = 317.8 mL
Therefore, the volume of oxygen produced when 1.00L of carbon dioxide reacts with 20.0g of Na2O2 is 317.8 mL, which is closer to your initial calculation of 1.57 L rather than the provided answer of 500 mL. It's possible that there was an error in the provided answer.
To determine the volume of oxygen produced, we need to use the ideal gas law equation:
PV = nRT
where:
P = pressure (in this case, 2.00 bar)
V = volume of gas (to be determined)
n = number of moles of gas (to be determined)
R = gas constant (given as 0.083145 L bar K-1 mol-1)
T = temperature (given as 25°C = 298 K)
First, let's calculate the number of moles of carbon dioxide (CO2) and sodium peroxide (Na2O2) involved in the reaction:
Molar mass of CO2 = 12.01 g/mol + 2(16.00 g/mol) = 44.01 g/mol
Number of moles of CO2 = mass / molar mass = 20.0 g / 44.01 g/mol ≈ 0.454 mol
Molar mass of Na2O2 = 2(22.99 g/mol) + 2(16.00 g/mol) = 77.98 g/mol
Number of moles of Na2O2 = mass / molar mass = 20.0 g / 77.98 g/mol ≈ 0.256 mol
Looking at the balanced reaction equation, we see that 1 mol of carbon dioxide reacts with 1 mol of sodium peroxide to produce 1 mol of oxygen. Therefore, the number of moles of oxygen produced is also 0.256 mol.
Now we can substitute the values into the ideal gas law equation:
PV = nRT
(2.00 bar)(V) = (0.256 mol)(0.083145 L bar K-1 mol-1)(298 K)
2.00V = 0.256(0.083145)(298)
2.00V ≈ 6.4639
V ≈ 6.4639 / 2.00
V ≈ 3.232 L
The volume of oxygen produced is approximately 3.232 L.
Therefore, it seems like there was an error in your calculation. The correct volume is approximately 3.232 L, not 1.57 L. It appears that the given answer of 500 mL (0.5 L) is incorrect.