Consider the following unbalanced equation: C2H5OH(g) + O2(g) → CO2(g) + H2O(l)

1.55 g of ethanol reacts with 10.0 g of oxygen. What is the total volume of gas present (in L) after the reaction is complete, assuming the reaction takes place at 1.00 atm and 25oC?
A) 0.572 L
B) 9.29 L
C) 4.11 L
D) 6.82 L
E) 1.65 L

D) 6.82 L

You will need to determine the limiting reagent, then calculate the volume of CO2 formed as well as the volume of the non-limiting reagent present at the end of the reaction. The apply PV = nRT and determine the total volume at the conditions listed. Post your work if you get stuck.

1.65

To determine the total volume of gas present after the reaction is complete, we first need to balance the chemical equation.

The unbalanced equation is:

C2H5OH(g) + O2(g) → CO2(g) + H2O(l)

To balance the equation, we want the same number of each type of atom on both sides of the equation.

Counting the number of each type of atom, we have:
Carbon (C): 2 on the left and 1 on the right
Hydrogen (H): 6 on the left and 2 on the right
Oxygen (O): 1 on the left and 3 on the right

To balance the equation, we can start by balancing the carbon atoms:
C2H5OH(g) + O2(g) → 2CO2(g) + H2O(l)

Next, balance the hydrogen atoms:
C2H5OH(g) + O2(g) → 2CO2(g) + 3H2O(l)

Finally, balance the oxygen atoms:
C2H5OH(g) + 3O2(g) → 2CO2(g) + 3H2O(l)

Now we can use the balanced equation to determine the moles of CO2 produced.

First, calculate the moles of ethanol (C2H5OH) using the given mass of ethanol and its molar mass. The molar mass of ethanol is calculated as follows:

C2H5OH:
(2 × 12.01 g/mol) + (6 × 1.01 g/mol) + 16.00 g/mol = 46.07 g/mol

The moles of ethanol (C2H5OH) can be calculated using the following equation:

moles = mass / molar mass
moles = 1.55 g / 46.07 g/mol ≈ 0.0337 mol

Next, calculate the moles of oxygen (O2) using the given mass of oxygen and its molar mass. The molar mass of oxygen is 32.00 g/mol. The moles of oxygen can be calculated as follows:

moles = mass / molar mass
moles = 10.0 g / 32.00 g/mol ≈ 0.3125 mol

From the balanced equation, we can see that the molar ratio between ethanol (C2H5OH) and carbon dioxide (CO2) is 1:2. Therefore, the moles of CO2 produced will be twice the moles of ethanol.

moles of CO2 = 2 × moles of ethanol
moles of CO2 = 2 × 0.0337 mol = 0.0674 mol

Now, we can use the ideal gas law to calculate the volume of gas at the given conditions of 1.00 atm and 25°C (298 K).

The ideal gas law equation is:

PV = nRT

Where:
P = pressure (in atm)
V = volume (in L)
n = moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature (in K)

We can rearrange the equation to solve for the volume:

V = (nRT) / P

Plugging in the values:

V = (0.0674 mol × 0.0821 L·atm/mol·K × 298 K) / 1.00 atm
V ≈ 1.577 L

Therefore, the total volume of gas present after the reaction is approximately 1.577 L.

However, the answer choices provided are given in 2 decimal places, so we need to round our answer. Rounding to 2 decimal places, the total volume of gas present is approximately 1.58 L.

Therefore, the correct answer is not listed among the provided answer choices.