I posted this question before and the answer is suppose to be 6.8 L but I keep getting 6.12 L
dinitrogen pentoxide n2o5 decomposes by a first-order decomposition in chloroform solvent to yield NO2 and O2. The rate constant at 45°C is 6.2 x 10^-4 mins^-1. Calculate the volume of O2 obtained from the reaction of 1.00 mol N2O5 at 45°C and 770 mmHg after 20.0 hours.
N2O5 = 2 NO2 + 1/2 O2
I think the problem is this.
I believe you have calculated N for N2O5 correctly. That's N2O5 LEFT. Then 1-N = mols N2O5 that decomposed to form O2, take 1/2 that and convert to V using PV = nRT
If I do that I obtain 6.76 which rounds to 6.8L. I don't remember but I may have omitted that 1-N step when I responded to your earlier post.
dinitrogen pentoxide n2o5 decomposes by a first-order decomposition in chloroform solvent to yield NO2 and O2. The rate constant at 45°C is 6.2 x 10^-4 mins^-1. Calculate the volume of O2 obtained from the reaction of 1.00 mol N2O5 at 45°C and 770 mmHg after 20.0 hours. Calculate the volume of 1 L of 0.600 M solution at 45°C
To calculate the volume of O2 obtained from the reaction of 1.00 mol N2O5, we need to first calculate the amount of O2 produced. We can then use the ideal gas law to convert the amount of O2 into the volume.
First, let's calculate the amount of O2 produced from 1.00 mol of N2O5.
According to the balanced chemical equation, 1 mole of N2O5 produces 1/2 mole of O2.
So, if we have 1.00 mol of N2O5, we will have 1.00 * (1/2) = 0.50 mol of O2.
Next, we can use the ideal gas law to convert the amount of O2 into volume.
The ideal gas law is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
To solve for volume (V), we can rearrange the equation as V = nRT/P.
Given:
n = 0.50 mol (amount of O2 produced)
R = 0.0821 L·atm/(mol·K) (ideal gas constant)
T = 45°C = 45 + 273 = 318 K (converted to Kelvin)
P = 770 mmHg = 770/760 atm (converted to atm)
Now, let's calculate the volume of O2:
V = (0.50 mol * 0.0821 L·atm/(mol·K) * 318 K) / (770/760 atm)
V = (0.5 * 0.0821 * 318) / (770/760)
V = 100.53 / 1.0064
V ≈ 100.12 L
Therefore, the volume of O2 obtained from the reaction of 1.00 mol N2O5 at 45°C and 770 mmHg after 20.0 hours is approximately 100.12 L.
It seems there might be a calculation or rounding error in your previous attempts, as the correct answer is approximately 100.12 L, not 6.8 L.
Please double-check your calculations and ensure you are using the correct values and units for all variables.