Acetylene Cylinder

Chemistry Helppppp!!!?

I need help with the following two questions:

1) Two cylinders of gas are used in welding. One cylinder is 1.2m high and 18 cm in diameter, containing oxygen gas at 2550 psi and 19 *C. The other is 0.76 m high and 28 cm in diameter, containing Acetylene Gas (ethyne) at 320 psi and 19 *C.
A) assuming complete combustion, which tank will empty first
B) What volume will the excess gas occupy?

2.) At 25 *C and 380 mm Hg, the density of sulfur dioxide is 1.31 g/L. The rate of effusion of sulfur dioxide through an orifice is 4.48 ml/s. What is the density of a sample of gas that effuses through an identical orifice at the rate of 6.78 ml/s under the same conditions? What is the molar mass of the gas?

PLEASE help me with these questions. I really can’t seem to figure them out. So please provide step by step instructions.

Thanks. =D

1.
The oxygen tank: diameter(d)=18cm, height(l)=120cm(1.2m).
You can calculate the tank’s volume:
V=((pi*d^2)/4)*l=((3.14*18^2)/4)*120=30536cm3=30.536dm3
The pressure is 2550psi. 1psi=6.894*10^3 Pa, so P=2550*6.894*10^3=17579kPa
Temperature is 19°C which is 292K( K=°C+273)
So you know the pressure, volume and temperature and you want to know many moles of oxygen you have. You can calculate it with the ideal gas law:
P*V=n*R*T, where R is the gas constant, R=8.314.
so n=(PV)/(RT)=221.112mol

The acetylene(C2H2) tank: You do the exact same calculations, like we did with the oxygen tank.
V=46.797dm3 P=2206kPa T=292K
PV=nRT –> n=42.523mol

Now that you have the amounts of gases, we can see which will we run out of first.
The combustion of ethyne is the following: 2C2H2 + 5O2 = 4CO2 + 2H2O
If we need 5 moles of O2 for 2 moles of C2H2 then we need 106.3moles of O2 for 42.523moles of C2H2. This means that the C2H2 tank will empty first, and we still have an excess of O2 gas.
The excess amount is 221.112mol-106.3mol=114.812mol of oxygen.
Now if we assume that the temperature did not change(292K) and the pressure in the O2 tank is the same as before(17579kPa), then we can calculate the O2′s volume with the ideal gas law (PV=nRT), this way V=15.855dm3

2.
T=25°C=298K
P=380mmHg=380torr=50.662kPa
Density(SO2)=1.31g/L=1.31g/dm3=0.00131g/cm3
M(SO2)=64g/mol
effusion rate=4.48ml/s=4.48cm3/s, this means that in one second 4.48cm3 SO2 gas leaves the system through the orifice. You can calculate the mass of the leaving gas, because you know it’s density and volume. V=4.48cm3, density=0.00131g/cm3, so m=0.0058688g which is (M=64g/mol) n=9.17*10^-5mol. So in one second 4.48cm3 SO2 gas leaves which is 9.17*10^-5moles.

The sample gas’s effusion rate is 6.78ml/s=6.78cm3/s. So in one second 6.78cm3 gas leaves.
you dont know the density or the molar mass, but you can still calculate it’s amount of substance(moles), because there environment has the same properties, meaning that the pressure and temperature is the same like at the SO2. This way we can use the ideal gas law: PV=nRT
we know P , T, R, V so n=1.386*10^-4. So in one second 6.78cm3 sample gas leaves which is 1.386*10^-4 moles.

This is as far as I can go, I dont know what to do from here. There is a law that says under the same circum stances this applies to 2 different gases:
density(1)/density(2)=molarmass(1)/molarmass(2). There are too many unknowns to calculate this equation.
Hope I helped!
Cheers!

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