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 7.00 l gas tank and a 4.00 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 155 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.

The pressure in the acetylene tank should be 108.5 atm

Explanation:

Via the ideal gas law, we can calculate the unknow number of pressure.

Step 1: The balanced equation

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

This means for 2 moles of acetylene consumed, there is needed 5 moles of oxygen to produced 4 moles of carbon dioxide and 2 moles of water.

Step2: Data given:

V1 = volume of tank 1 = 7L

V2 = volume of tank 2 = 4L

P1 = pressure of tank 1 = 155 atm

P2= pressure of tank 2 = TO BE DETERMINED

n1 = moles of the oxygen tank = 5 moles

n2 = moles of the acetylene tank = 2 moles

Step 3: Calculate the pressure of the acetylene tank.

The ideal gas law says:

(P1*V1)/n1 = (P2*V2)/n2

To find the pressure P2 of the acetylene tank: P2 = (P1*V1 *n2)/ (n1*V2)

P2 = (155 atm * 7L *2moles)/(5moles * 4L) = 108.5 atm

The pressure in the acetylene tank should be 108.5 atm

the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

i have no clue hope you find the answer