A rigid 5.00 L cylinder contains 24.5g of N2(g) and 28.0g of O2(g).
(a). Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K
(b) The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the following.
1) The mole function of N2(g) in the cylinder
2) The partial pressure, in atm, of N2(g) in the cylinder
(c) if the cylinder develops a pihole-sized leak and some of the gaseous mixture escapes, would the ratio
moles of n2(g)/Moles of O2(g) in the cylinder increase, decrease, or remain the same? Justify your answer
A rigid 5.00 L cylinder contains 24.5g of N2(g) and 28.0g of O2(g).
Convert 24.5 g N2 and 28 g O2 to mols. Remember mols = grams/molar mass.
(a). Calculate the total pressure, in atm, of the gas mixture in the cylinder at 298 K
Use the number of mols of each gas and PV=nRT to calculate p for N2 and do the same to calculate p for O2. Then the total P = pN2 + pO2.
(b) The temperature of the gas mixture in the cylinder is decreased to 280 K. Calculate each of the following.
1) The mole function of N2(g) in the cylinder I'm sure you intended to write mole fraction of N2. Mole fraction N2 = mols N2/total number of mols. All of this information is available from the work at the beginning.
2) The partial pressure, in atm, of N2(g) in the cylinder
Calculate total pressure in the cylinder. Use PV=nRT, use the total number of mols N2 + O2. Don't forget to use the new temperature of 280 K. Then pN2 = total P*mole fraction N2.
(c) if the cylinder develops a pihole-sized leak and some of the gaseous mixture escapes, would the ratio
moles of n2(g)/Moles of O2(g) in the cylinder increase, decrease, or remain the same? Justify your answer
Read about Graham's Law of diffusion (or effusion)depending upon which book you are using. Applying that law will tell you what will happen to the ratio.
N2 converted is 0.874molN2 and O2 is 0.875molO2
N2: P=? PV=nRT
V=5.ool PV/V=nRT/V
n=0.874molN2 P=nRT/V
R=0.21latm/molk
T=298k
P=0.874+0.821+298/5.00=299.695/5.oo=
59.9392
O2: P=? PV=nRt
V=5.ool PV/V=nRT/V
n=0.875molO2 P=nRT/V
R=0.821latm/molk
T=298k
P=0.875+0.21+298/5.00=299.696/5.00=
59.9392
P=59.939O2+59.939N2=119.88
so for the next part, does it mean that N2 is 0.874molN2
2) P=?
V=5.00L
n=1.749molN2O2
R=0.821latm/molk
T=280k
P=1.749+0.821+280/5.00+282.57/5.00
P=56.514
PN2=56.514*0.874
PN2=49.39
Are my answers correct, did i do it right?
No because when looking for the pressure of N2 and O2 you added the values of .0821atm/molk & 280K & .875mol instead of multiplying then
A different rigid 5.00L cylender contains .176 mol of NO(g) at 298K. A .176 mol sampl eof O2(g) is added to the cylender , where a reaction occurs of produce NO2(g).
d. Write the balanced equation for the reaction.
e. Calculate the total pressure, in atm, in the cylender at 298K after the reaction is complete
D. 2NO + O2 -> 2NO2
E. since only half of the moles of O2 will be consumed in the reaction there will be .176 mol NO2 and .088 mol O2
so...
PNO2 = .176mol(.08296)(298K)/5L = .8608 atm
PO2 = .088mol(.08206)(298K)/5L = .4304 atm
PNO2 + PO2 = Total Pressure = 1.2912 atm
5 b) (chi)x1= n1/ n total = N1/ n1 + n2
x1 = .875/.875+.875=.5 the molar fraction
PARTIAL PRESSURE is (x)n2*P total= partial pressure of N2 \
.5 *8.56atm = 4.28atm
A solution of hydrogen peroxide, H2O2 , is titrated with potassium permanganate, KMnO4 , according to the following equation:
5 H2O2 + 2 KMnO4+ 3H2SO4 -> 5O2 + 2 MnSO4 +8H2O + K2SO4
It requires 46.9mL of 0.145 mol/L KMnO4 to titrate 50.0 mL of the solution of H2O2 .
What is the moles per litre concentration of H2O2 in the solution?
For R you need to use .08206 or .0821
And for Pv=nRT You just need to do it as is..
To write the balanced equation for the reaction between NO and O2, we need to determine the products formed. The reaction can be represented as:
2NO + O2 -> 2NO2
Now, let's calculate the total pressure in the cylinder at 298K after the reaction is complete.
First, we need to find the partial pressure of NO and O2 in the cylinder before the reaction. Since there are 0.176 mol of NO and 0 mol of O2 initially, the partial pressure of NO is determined using the ideal gas law:
PV = nRT
P(NO) = (0.176 mol) * (0.0821 atm/molK) * (298 K) / (5.00 L) = 0.0924 atm
The partial pressure of O2 is 0 atm since there is no O2 initially.
Now, after the reaction is complete, all NO reacts to form NO2. Since there are 0.176 mol of NO initially, there will be 0 mol of NO remaining. However, the same amount of NO2 is produced as the stoichiometry of the balanced equation shows. Therefore, there will be 0.176 mol of NO2 in the cylinder.
To calculate the total pressure after the reaction, we need to find the partial pressure of NO2:
P(NO2) = (0.176 mol) * (0.0821 atm/molK) * (298 K) / (5.00 L) = 0.0924 atm
Therefore, the total pressure in the cylinder after the reaction is complete is the sum of the partial pressures of NO2 and O2 (which is now present due to the reaction):
Total pressure = P(NO2) + P(O2)
Total pressure = 0.0924 atm + 0 atm = 0.0924 atm
So, the total pressure in the cylinder at 298K after the reaction is complete is 0.0924 atm.