92.01 grams of N2O4 (g) is placed in a container and allowed to dissociate.
N2O4 (g) --> 2NO2 (g)
The mixture of N2O4 and NO2 resulting from the reaction occupies 36.0 liters
at a total pressure of 773 mmHg and 45 °C.
A. Let x equal the moles of N2O4 that dissociate. How many moles of N2O4
remain un-dissociated, and how many moles of NO2 are produced?
B. What are the mole fractions of N2O4 and NO2 in the mixture?
C. What are the partial pressures of N2O4 and NO2 in the mixture?
initial moles N2O4 = grams/molar mass = 92.01/92.01 = 1.00 moles.
.............N2O4 ==> 2NO2
initial.......1.0......0
change.........-x......2x
equil.......1-x........2x
and that answers #1. I don't know if you want a number or not but that can be obtained after you finish the rest of the problem.
B. Total moles = 1-x+2x = 1+x
moles N2O4 = 1-x
moles NO2 = 2x
XN2O4 = (1-x)/(1+x)
XNO2 = 2x/(1+x)
Again, if a number is required, you can finish at the end.
C. PN2O4 = XN2O4*773/760
PNO2 = XNO2*773/760
Do you want numbers? If so, then use PV = nRT and solve for n. Using total volume, P, and T, you end up with total n. I solved this and came up with approximately 1.2 but you need to do it more accurately than that. Then you know total moles is 1+x from above. Since 1+x = 1.2, you can solve for x. Knowing x allows you to solve for mole fractions in B and partial pressure in C.
VAPOUER DENSITY OF MIXTURE OF N2O4&NO2 is 38.3 at 25degree caliculate number of NO2 MOLICULES IN100 MOLES MIXTURE
To solve this problem, we need to use the ideal gas law (PV = nRT) to find the number of moles of N2O4 and NO2, as well as the partial pressures of N2O4 and NO2 in the mixture. Here's how to do it step by step:
A. Let's assume that x is the number of moles of N2O4 that dissociate. Since 1 mole of N2O4 produces 2 moles of NO2, we can conclude that 2x moles of NO2 will be produced.
To find the number of moles of N2O4 that remain un-dissociated, subtract the moles of N2O4 that dissociate (x) from the initial moles of N2O4 (92.01 grams). To do this, we need to convert grams of N2O4 to moles using its molar mass.
1. Find the molar mass of N2O4:
Molar mass of N = 14.01 g/mol
Molar mass of O = 16.00 g/mol
Total molar mass of N2O4 = (2 * 14.01) + (4 * 16.00) = 92.02 g/mol (approximately)
2. Convert the mass of N2O4 to moles:
Moles of N2O4 = Mass of N2O4 / Molar mass of N2O4
= 92.01 g / 92.02 g/mol (approximately)
≈ 1.000 moles of N2O4
Now, since 1 mole of N2O4 produces 2 moles of NO2:
Moles of NO2 = 2 * x
Moles of N2O4 un-dissociated = 1.000 moles - x
B. The mole fraction of a component in a mixture can be calculated as the moles of that component divided by the total moles of all components.
Mole fraction of N2O4 = Moles of N2O4 / Total moles of N2O4 and NO2
= Moles of N2O4 / (Moles of N2O4 + Moles of NO2)
= Moles of N2O4 / (1.000 moles + 2x)
Mole fraction of NO2 = Moles of NO2 / Total moles of N2O4 and NO2
= Moles of NO2 / (Moles of N2O4 + Moles of NO2)
= 2x / (1.000 moles + 2x)
C. To find the partial pressures of N2O4 and NO2, we can use Dalton's Law of partial pressures. According to the law, the total pressure exerted by a mixture of ideal gases is equal to the sum of the partial pressures of the individual gases.
Partial Pressure of N2O4 = Mole fraction of N2O4 * Total pressure
= Mole fraction of N2O4 * 773 mmHg
Partial Pressure of NO2 = Mole fraction of NO2 * Total pressure
= Mole fraction of NO2 * 773 mmHg
That's how you calculate the moles of N2O4 and NO2, the mole fractions of N2O4 and NO2, and the partial pressures of N2O4 and NO2 in the mixture.