Question

Acetylene torches are used for welding. These torches use a mixture of acetylene gas, C2H2 , and oxygen gas, O2 to produce the following combustion reaction: 2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g) Part A Imagine that you have a 6.50 L gas tank and a 4.50 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 145 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.

Answer 1

Answer: 80.14

Step-by-step explanation:

The first step to solve this problem is to calculate the amount of oxygen gas needed to react with the given amount of acetylene gas. According to the balanced equation of the combustion reaction, 5 moles of O2 are needed for every 2 moles of C2H2:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

Thus, we can use the following proportion to calculate the amount of oxygen gas needed:

5 mol O2 / 2 mol C2H2 = x mol O2 / y mol C2H2

where x and y are the numbers of moles of oxygen and acetylene gases, respectively, needed to fill each tank.

To calculate the pressure needed to fill the acetylene tank, we can use the ideal gas law:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Assuming that the temperature and volume are constant for both tanks, we can write:

P1 / P2 = n1 / n2

where P1 and P2 are the pressures in the oxygen and acetylene tanks, respectively, and n1 and n2 are the number of moles of each gas.

Now, we can combine these two equations to solve for P2, the pressure in the acetylene tank:

5 / 2 = x / y

y = 2x / 5

P1 / P2 = n1 / n2 = (5 mol / 4.50 L) / (2x / 6.50 L)

P2 = P1 * (2x / 5) * (4.50 L / 6.50 L)

P2 = 145 atm * (2/5) * (4.50/6.50)

P2 = 80.14 atm

Therefore, the acetylene tank should be filled to a pressure of 80.14 atm to ensure that both tanks run out of gas at the same time.