Ethanol burns to produce 1300.0 kJ/mol and has a heat capacity of 65.6 J/mol·K. If Dr. Porter has 1.00 L of liquid ethanol (density = 0.789 g/mL), what is the minimum mass of fuel that should be removed and burned to heat the remaining ethanol from 25 °C to 90 °C? (Assume the equilibrium occurs in the gas phase.) Hint: Set-up an equation with the unknown X = ‘mass of ethanol’, where when units are canceled, you’re left with Joules = Joules.

I would do this although it may be the long way around. I would convert all of the "moles" to grams.

1300 kJ/mol = 1300000/46.069 = ??J/g for the heat of combustion.
65.6 J/mol = 65.6/46.00 = xx J/g for the heat capacity (specific heat) of ethanol.
1.00 L ethanol = 1000 mL and that has a mass of 1000 x 0.789 g/mL = xx grams.

Let X = mass ethanol removed.
Xg*heat combustion = (mass ethanol-X)*specific heat ethanol* (Tfinal-Tinitial).
Solve for X and 789-X.
I obtain something a little less than 3 grams to be removed and that will heat the remainder from 25 to 90.
Post your work if you get stuck.

Again Melissa?

To find the minimum mass of fuel that should be burned to heat the remaining ethanol from 25 °C to 90 °C, we can use the equation:

q = m * C * ΔT

Where:
q = heat energy required (in Joules)
m = mass of ethanol (in grams)
C = heat capacity of ethanol (in J/mol·K)
ΔT = change in temperature (in K)

First, we need to convert the volume of ethanol to mass using the density.

Given:
Volume of ethanol = 1.00 L
Density of ethanol = 0.789 g/mL

Mass of ethanol = volume × density
Mass of ethanol = 1.00 L × 0.789 g/mL = 0.789 g

Now, we can calculate the heat energy required using the equation and given values.

ΔT = (90 °C - 25 °C) = 65 K

q = m * C * ΔT
q = 0.789 g * 65.6 J/mol·K * 65 K

Since we are given the heat of combustion for ethanol, we can find the number of moles of ethanol corresponding to 1300.0 kJ/mol.

1300.0 kJ/mol = q
1300.0 kJ/mol = 0.789 g * 65.6 J/mol·K * 65 K

Now, we can rearrange the equation to solve for the mass of ethanol (X):

X = q / (65.6 J/mol·K * 65 K)

Substitute the value of q:

X = 1300.0 kJ/mol / (65.6 J/mol·K * 65 K)

Convert kJ to J:

X = (1300.0 × 1000) J/mol / (65.6 J/mol·K * 65 K)

Calculate the minimum mass of ethanol:

X = 2.00 mol

Finally, convert moles to grams using the molar mass of ethanol:

Molar mass of ethanol (C2H5OH) = (2 * C) + (6 * H) + (1 * O)
Molar mass of ethanol = (2 * 12.01 g/mol) + (6 * 1.01 g/mol) + (1 * 16.00 g/mol) = 46.07 g/mol

Minimum mass of ethanol = X * molar mass of ethanol
Minimum mass of ethanol = 2.00 mol * 46.07 g/mol = 92.14 g

Therefore, the minimum mass of fuel that should be removed and burned to heat the remaining ethanol from 25 °C to 90 °C is 92.14 grams.

To find the minimum mass of fuel that should be removed and burned to heat the remaining ethanol, we need to use the equation "Joules = Joules".

First, we need to calculate the initial heat energy (Joules) of the liquid ethanol. We know that ethanol has a heat capacity of 65.6 J/mol·K.

Step 1: Convert volume to mass
Given that the density of liquid ethanol is 0.789 g/mL and Dr. Porter has 1.00 L of ethanol, we can convert the volume to mass as follows:

1.00 L * 0.789 g/mL = 789 g

So, Dr. Porter has 789 grams of ethanol.

Step 2: Convert mass to moles
To convert mass (in grams) to moles, we need to divide the mass by the molar mass of ethanol. The molar mass of ethanol is approximately 46.07 g/mol.

789 g / 46.07 g/mol ≈ 17.12 mol

Now we know that Dr. Porter has approximately 17.12 moles of ethanol.

Step 3: Calculate initial heat energy
The initial heat energy (Joules) can be calculated using the formula:

Initial heat energy = Heat capacity * moles * temperature difference

Given:
Heat capacity of ethanol: 65.6 J/mol·K
Temperature difference: 90 °C - 25 °C = 65 °C

Initial heat energy = 65.6 J/mol·K * 17.12 mol * 65 °C

Step 4: Calculate the minimum mass of fuel
Assuming the ethanol reaches equilibrium in the gas phase and is completely burned, the energy released by burning the fuel is 1300.0 kJ/mol.

So, the minimum mass of fuel (in moles) needed can be calculated using the formula:

Energy released by fuel burn = Heat capacity * moles * temperature difference

1300.0 kJ/mol = 65.6 J/mol·K * moles * 65 °C

Solving for moles:

moles = 1300.0 kJ/mol / (65.6 J/mol·K * 65 °C)

moles ≈ 3.46 mol

Step 5: Convert moles to mass
To convert moles back to mass, we multiply by the molar mass of ethanol:

Mass of fuel = moles * molar mass

Mass of fuel = 3.46 mol * 46.07 g/mol

Mass of fuel ≈ 159.36 g

Therefore, the minimum mass of fuel that should be removed and burned to heat the remaining ethanol from 25 °C to 90 °C is approximately 159.36 grams.