GenCHEM Help PLEASE!!!!

[Dion]

--- A 3.31-g sample of lead nitrate is heated in an evacuated cylinder with a volume of 1.62 L. The salt decomposes when heated, according to the equation shown below.

2 Pb(NO3)2 (s) 2 PbO (s) + 4 NO2 (g) + O2 (g)

Assuming complete decomposition of the lead nitrate, what is the total gas pressure inside the cylinder after decomposition and cooling to a temperature of 300 K? Assume the PbO (s) takes up negligible volume.

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--- Air is 20% oxygen and 80% nitrogen. What is the mass of air in an automobile tire of 19.7 L and internal pressure of 46.7 PSI at 297 K? (That pressure is the same as the 32 PSI difference you usually measure as the tire pressure 32 PSI + 14.7 PSI.; 14.7 PSI = 1 atm)

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--- How many liters of oxygen at 730 torr and 300 K is needed to burn one kilogram of isopropyl alcohol, C3H7OH?
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---A sample of N2 gas is contaminated with a gas (A) of unknown molar mass. The partial pressure of each gas is known to be 200 torr at 298 K The gases are allowed to effuse through a pinhole, and it is found that gas A escapes at three times the rate of N2. Calculate the molar mass of gas A. Suggest a possible structure.
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[Lily-VA]

um..can't believe how these Q's are!
for the 2nd Q...I would do...

n=19.7L*32PSI /(8.31* 297K :constant..don't remember)
Then...l don't know. Time to go to bed..my brain can't think. Will try again tomorrow..if I come up with something..then...

 

[erichen]

--- A 3.31-g sample of lead nitrate is heated in an evacuated cylinder with a volume of 1.62 L. The salt decomposes when heated, according to the equation shown below.

2 Pb(NO3)2 (s) 2 PbO (s) + 4 NO2 (g) + O2 (g)

Assuming complete decomposition of the lead nitrate, what is the total gas pressure inside the cylinder after decomposition and cooling to a temperature of 300 K? Assume the PbO (s) takes up negligible volume.

------------------------------------------------------------------------

--- Air is 20% oxygen and 80% nitrogen. What is the mass of air in an automobile tire of 19.7 L and internal pressure of 46.7 PSI at 297 K? (That pressure is the same as the 32 PSI difference you usually measure as the tire pressure 32 PSI + 14.7 PSI.; 14.7 PSI = 1 atm)

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--- How many liters of oxygen at 730 torr and 300 K is needed to burn one kilogram of isopropyl alcohol, C3H7OH?
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---A sample of N2 gas is contaminated with a gas (A) of unknown molar mass. The partial pressure of each gas is known to be 200 torr at 298 K The gases are allowed to effuse through a pinhole, and it is found that gas A escapes at three times the rate of N2. Calculate the molar mass of gas A. Suggest a possible structure.
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i believe that number one is easier than it looks. if all these gases are treated as ideal gases, then as long as we know how many moles of each gas were created, we can add them up as if they all are one homogenous gas. *in ideal gases, 1 mole of gas occupies 22.4L@ STP, but notice that 22.4 L can't be used in this problem because the system is not @ STP*

so first we need to find how many moles of each gas were produced.
We are given 3.31g of Pb(NO3)2, which has a molar mass of (207.2g)

(3.31g)/(207.2g/mol)x(4 mol NO2/2 mol Pb(NO3)2) <-molar ratio
which yields: 0.031949806 mol NO2 gas

likewise, find the number of moles of 02

(3.31g)/(207.2g/mol)x(1mol 02/ 2 mol Pb(NO3)2)
which yields: 0.007987 mol O2 gas

since we are going to treat them like a homogenous gas, we can add the number of moles together.

0.031959806 mol NO2 + 0.007897 mol 02= 0.039936806 mol NO2 and O2 gas combined.

then use the ideal gas equation to find the pressure.

PV=nRT
P= nRT/V
P= ((0.039936806 mol gas)(0.08206 <-ideal gas constant)(300K)) / (1.63L)
P=0.603168276 atm

i hope this is correct 🙂

 

[doc toothache]

The molecular weight of lead nitrate is, conveniently, 331.23.

moles of lead nitrate=~ 3.31 /331=0.01
since 2 moles of lead nitrate yields 5 moles of gas, then
2/0.01=5/x; then x=0.025 moles of gas (moles of gas at STP)
Since 1 mole occupies 22.4 L; then 0.025moles would occupy 0.56L

V1P1T2=V2P2T1
(0.56)(1)(300)=(1.62)(P2)(298)
P=0.38atm

2 C3H7OH +9 O2 = 6 CO2 + 8 H2O
moles of isopropyl alc= 1000/60=16.6
2/9=16.6/x x=74.7 moles of oxygen needed
since 1 moles of O2 occupies 22.4 L; total needed 1673 L

V1P1T2=V2P2T1; V=1673(1)(300)/0.96x298
V=1.755kL

 

[erichen]

sorry, i didn't count in the NO3 for the molec weight of PbNO3

im so stupid 🙁.

but doc, its not at STP. 22.4L/mole of gas is only for ideal gases at STP is it not?

 

[doc toothache]

?? The conditions are not ideal. P1V1T2=P2V2T1 was used to convert to the stated conditions.??

 

[doc toothache]

Convert psi into atm 46.7/14.7=3.18 atm
Part P O2= (3.18)(.20)=.636 atm
Part P N2= (3.18)(.80)=2.54 atm

for O2 PV=nRT
(0.636)(19.7)=n(0.08)(297), solving for n, n=0.53 moles
g O2= (moles O2)(mol. wgt O2), g O2=16.96

for N2 PV=nRT
(2.54)(19.7)=n(0.08)(297), solving for n, n=2.10 moles
g N2=(moles N2)(mol wgt N2), g=58.8
total weight= 75.76 g

 

[Lily-VA]

I got the matching answers above. Good job.
I still don't know the effusion rate one. Didn't do much in class...🙂

 

[Leh]

once u have deduced the numbers of moles (0.025) u can also plug .025 (total moles of gas) into the ideal gas law PV=nRT to get the pressure....p(1.62L)=(.025moles)(.08206)(300) to get the same answer...
the partial pressure law gives the same answer too.

 

[doc toothache]

R1/R2= square root of (M 2/M 1)

3/1= square root(28/x)
x has a molecular weight of 3.11.
Gas is 3 He (a radioisotope of helium)

 

[Dion]

Thanks Guys!

 

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