[s133kgtvr4]

I have two problems that I am not sure how to do. if someone can explain to me how to solve it, that would be great.

1)Imagine that you have a 7.00 L gas tank and a 2.50 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 atm, to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time?


2)A gaseous mixture of O_2 and  N_2 contains 30.8 % nitrogen by mass. What is the partial pressure of oxygen in the mixture if the total pressure is 505 mmHg?

[Borek]

1. Start with the balanced reation equation.

2. Assuming 100g of mixture - how many moles of oxygen? Of nitrogen?

[brianfreytag]

The first one is pretty easy.  It is simply one of the gas laws.
All you need to know is the combined gas law:  (P1V1)/T1=(P2V2)/T2.  Or more simply P1V1T2=P2V2T1

For this, ignore the temperature.  You know P1 and V1, and V2.  So plug in the numbers.

P2= (P1V1)/V2

[vkut79]

So you assume that you need a 1:1 ratio of O2 moles to Acetylene moles for the welding reaction? Its not mentioned in the problem explicitly.

[LQ43]

The first one is pretty easy.  It is simply one of the gas laws.
All you need to know is the combined gas law:  (P1V1)/T1=(P2V2)/T2.  Or more simply P1V1T2=P2V2T1

For this, ignore the temperature.  You know P1 and V1, and V2.  So plug in the numbers.

P2= (P1V1)/V2

No. Please READ the problem, there are 2 compounds in a reaction and so need a balanced equation as Borek has suggested.

Using the Combined Gas law is not appropriate here. The temperature cannot be ignored.

[Borek]

No. Please READ the problem, there are 2 compounds in a reaction and so need a balanced equation as Borek has suggested.

Using the Combined Gas law is not appropriate here. The temperature cannot be ignored.

I think you are both wrong - you need the balanced equation and you can ignore the temperature (more precisely - both gases have the same temperature, so it cancels out)

[pcxft]

for question 1)

If we assume complete combustion, we get only CO2 and H2O as products. No CO for instance. Hence, the equilibrated reaction is:

2.C2H2 + 5.O2 ----->  4.CO2 + 2.H2O

We need 5 moles of oxygen for 2 moles of acetylene:

n(A) / 2 = n(O) / 5  which is equivalent to  n(O) / n(A) = 5 / 2
where n(A) is the mole number of acetylene and n(O) the mole number of oxygen.

To finish this problem, we apply the gas law for both tanks:
P(O) x V(O) = n(O) x R x T    and     P(A) x V(A) = n(A) x R x T

If we assume both tanks are at the same temperature, then we can divide one equation by the other one:
P(O)V(O) / P(A)V(A) = n(O) / n(A)   
R and T cancel out

but, we saw that  n(O) / n(A) = 5/2,  P(O)=115atm,  V(O)=7L  and V(A)=2.5L

then   (115 x 7) / (P(A) x 2.5) = 5/2

P(A) = 128.8atm     Hope my explanation is clear and without mistake

[Borek]

Hope my explanation is clear and without mistake

Hope you will read forum rules before solving other people homework.