Question

A 22.5 mL sample of liquid ethanol (C₂H₅OH, density = 0.789 g/mL) was injected into a 10.0 L cylinder containing O₂ at a pressure of 4.50 atm and a temperature of 25°C. The cylinder was heated to 125°C and a spark was used to ignite the ethanol, which was completely combusted. What was the final total pressure in the cylinder (in atm) after the reaction, while it was still at 125°C?

Answer 1

Answer:

The final total pressure in the cylinder = 8.53 atm

Explanation:

From the information given:

The volume of liquid ethanol = 22.5 mL

The density of the liquid ethanol = 0.789 g/mL

The volume of the cylinder = 10.0 L

The pressure of the cylinder = 4.50 atm

The temperature of the cylinder = 25°C = (273.15 + 25) K = 298.15 K

Using the ideal gas equation:

PV = nRT

where n is the number of moles before the reaction:

$n = \dfrac{PV}{RT}$

$n = \dfrac{4.5 \ \times 10 }{0.08205 \times 298.15}$

n = 1.8395 moles

Recall that:

Density = mass /volume

Thus; the mass of liquid ethanol = (Density × volume ) of liquid ethanol

the mass of liquid ethanol = 0.789g/mL  × 22.5 mL

the mass of liquid ethanol = 17.7525 g

Since we know the mass of the liquid ethanol =  17.7525 g and the standard molar mass of liquid ethanol = 46.07 g/mol

Then:

the number of moles of liquid ethanol = 17.7525 g / 46.07 g/mol

the number of moles of liquid ethanol = 0.38534 mol

The chemical equation for the reaction can be represented as:

$C_2H_5OH + 3O_2 \to 2CO_{2(g)} + 3H_2O_{(g)}$

From above 1 mole of ethanol react with 3 moles of oxygen to produce 2 moles of carbon dioxide and 3 moles of water

Since each mole of ethanol reacts with 3 moles of oxygen

Then:

0.38534 mol of ethanol = (0.38534 mol × 3) oxygen

Therefore, the amount of oxygen molecule left in the cylinder after the reaction = 1.8395 mol - (0.38534 mol × 3)

= 1.8395 mol - 1.15602 mol

= 0.68348 mol

The number of moles of carbon dioxide formed = 2 × moles of ethanol

number of moles of carbon dioxide formed =  2 × 0.38534 mol

number of moles of carbon dioxide formed =   0.77068 mol

The number of moles of water vapour formed = 3 × 0.38534 mol

The number of moles of water vapour formed = 1.15602 mol

∴

The total moles of gas = moles of oxygen left + moles of water vapour + moles of carbon dioxide

The total moles of gas =  0.68348 mol + 0.77068 mol + 1.15602 mol

The total moles of gas = 2.61018 mol

Finally, to find the total pressure in the cylinder when the temperature is 125°C = (273.15 + 125)K

= 398.15 K

Then; using the ideal gas equation PV = nRT

P = nRT/V

P = ( 2.61018  × 0.08205 × 398.15 ) / 10

P = 85.2699/10

P = 8.52699 atm

P ≅ 8.53 atm