Question

Imagine that you have a 5.50 L gas tank and a 2.00 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 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.

Butane, C4H10

, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)

At 1.00 atm

and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 4.00 g of butane?

And, Butane, C4H10

, is a component of natural gas that is used as fuel for cigarette lighters. The balanced equation of the complete combustion of butane is

2C4H10(g)+13O2(g)→8CO2(g)+10H2O(l)

At 1.00 atm

and 23 ∘C, what is the volume of carbon dioxide formed by the combustion of 4.00 g of butane?

Answer 1

Answer:-

(I)-

Given:-

volume of oxygen (V1) = 5.50 L

pressure of oxygen (P1) = 145 atm

volume of acetylene (V2) =  2.00 L

pressure in the acetylene tank (P2) = ?

Since the at constant temperature

P1V1 = P2V2

So

pressure of oxygen (P1)volume of oxygen (V1) = pressure in the acetylene tank (P2)volume of acetylene (V2)

145 atm   5.50 L = pressure in the acetylene tank (P2) 2.00 L

pressure in the acetylene tank (P2) = 145 atm   5.50 L / 2.00 L

pressure in the acetylene tank (P2) = 398.75 atm

but we know that

2C2H2(g) + 5O2(g)      4CO2(g)   + 2H2O(l)

2 moles 5 moles 4 moles 2 moles

As mentioned in above equation 2 moles of acetylene (C2H2) reacts with 5 moles of oxygen (O2) to form CO2 and water (H2O) therefore same no. of moles of acetylene (C2H2) and oxygen (O2) do not required for welding torch. It would be 2 / 5 of pressure in the acetylene tank (P2).

then

pressure of acetylene tank to ensure that you run out of each gas at the same time (PC2H2) = 2 / 5 pressure in the acetylene tank (P2).

pressure of acetylene tank to ensure that you run out of each gas at the same time (PC2H2) = 2 / 5 398.75 atm

pressure of acetylene tank to ensure that you run out of each gas at the same time (PC2H2) = 797.5 / 5 atm

pressure of acetylene tank to ensure that you run out of each gas at the same time (PC2H2) = 159.5 atm (i.e the answer)

(II)-

Given:-

wt. of butane (C4H10) = 4.00 g

pressure (P) = 1.00 atm

temperature (T) = 23 0C = 23 + 273 = 296 K

gas constant (R) = 0.08206 L.atm.K-1.mol-1

volume of carbon dioxide (CO2) = ?

Since

molar mass of butane (C4H10) = 4 molar mass of C + 10 molar mass of H

molar mass of butane (C4H10) = 4 12 + 10 1

molar mass of butane (C4H10) = 48 + 10

molar mass of butane (C4H10) = 58 g/mol

therefore

no. of moles butane (nC4H10) = wt. of butane (C4H10) / molar mass of butane (C4H10)

no. of moles butane (nC4H10) = 4.00 g / 58 g/mol

no. of moles butane (nC4H10) = 0.0690 mol

As we know that

2C4H10(g) + 13O2(g)      8CO2(g)   + 10H2O(l)

So

2 moles of butane (nC4H10) completely reacts with O2 to produce = 8 moles of CO2

1 moles of butane (nC4H10) completely reacts with O2 to produce = 8 / 2 moles of CO2​​​​​​​

then

0.0690 moles of butane (nC4H10) completely reacts with O2 to produce = 8 0.0690 / 2 moles of CO2​​​​​​​

0.0690 moles of butane (nC4H10) completely reacts with O2 to produce = 0.276‬ moles of CO2​​​​​​​

therefore

no. of moles carbon dioxide (nCO2) = 0.276‬ mol

So according to the formula

PV = nRT

V = nRT / P

then

volume of carbon dioxide (VCO2) = nCO2RT / P

volume of carbon dioxide (VCO2) = 0.276 mol 0.08206 L.atm.K-1.mol-1 296 K / 1.00 atm

volume of carbon dioxide (VCO2) = 6.70 L (i.e the answer)