1.86 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?
convert the reaction to moles.
each mole occupies 22.4L at STP, so adjust that volume using PV=kT
you balanced this wrong lol
Well, that sounds like quite the explosive mixture! But don't worry, I'm here to help and make it fun. Let's break it down.
First, we need to find the number of moles of ethanol and oxygen. To do that, we divide the given masses by their respective molar masses.
The molar mass of ethanol (C₂H₅OH) is approximately 46 g/mol, so 1.86 g of ethanol is about 0.04 moles.
For oxygen (O₂), the molar mass is 32 g/mol, so 10.0 g of oxygen is roughly 0.31 moles.
Now, let's check the balanced equation for this reaction.
C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O
From the balanced equation, we can see that for every 1 mole of ethanol, 3 moles of oxygen are required.
Since we have 0.04 moles of ethanol, we need 3 times that amount of oxygen, which is 0.12 moles.
Now, let's assume the reaction goes to completion, which means all the ethanol and oxygen react and no gas is left unreacted.
In total, we have 0.04 moles of ethanol and 0.12 moles of oxygen, giving us a total of 0.16 moles of gas.
To find the total volume, we can use the ideal gas law, which states that PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant (0.0821 L•atm/mol•K), and T is the temperature in Kelvin.
We know the pressure is 1.00 atm and the temperature is 25°C, which is 298 K.
Plugging in the values, we get:
(1.00 atm) * V = (0.16 moles) * (0.0821 L•atm/mol•K) * (298 K)
Simplifying, we find:
V = (0.16 moles) * (0.0821 L•atm/mol•K) * (298 K) / (1.00 atm)
Doing the math, we get approximately 3.94 L.
So, after the reaction is complete, we'll have a total volume of gas of around 3.94 Liters. That's a lot of gas, so make sure you have good ventilation!
To find the total volume of gas present after the reaction is complete, we need to determine the moles of each reactant and then use the ideal gas law equation.
1. First, we calculate the moles of ethanol and oxygen using their respective molar masses.
- Molar mass of ethanol (C2H5OH) = 46.07 g/mol
- Moles of ethanol (C2H5OH) = mass / molar mass = 1.86 g / 46.07 g/mol
2. Repeat the same procedure for oxygen:
- Molar mass of oxygen (O2) = 32 g/mol
- Moles of oxygen (O2) = mass / molar mass = 10.0 g / 32 g/mol
3. The balanced chemical equation for the reaction of ethanol with oxygen is:
C2H5OH + 3O2 → 2CO2 + 3H2O
We can see from the equation that for every 1 mole of ethanol that reacts, 3 moles of oxygen are consumed.
4. From the moles of ethanol and oxygen, we can determine which reactant is limiting. The reactant that produces fewer moles will be the limiting reactant.
Divide the moles of each reactant by their stoichiometric coefficients to determine the number of moles required for the reaction:
- Ethanol: 1.86 g / 46.07 g/mol = 0.040 mol
- Oxygen: 10.0 g / 32 g/mol = 0.31 mol
Since we need 3 moles of oxygen for every mole of ethanol, we see that ethanol is the limiting reactant since it produces fewer moles. Therefore, we can use the moles of ethanol to calculate the moles of carbon dioxide produced.
5. Now, we can use the ideal gas law equation, PV = nRT, to calculate the volume of gas produced. Rearrange the equation to isolate V (volume):
V = nRT / P
Where:
- V is the volume of gas (in liters)
- n is the moles of gas
- R is the ideal gas constant (0.0821 L·atm/mol·K)
- T is the temperature in Kelvin
- P is the pressure in atmospheres
We need to be consistent with units, so let's convert everything to Kelvin:
- Temperature in Kelvin = 25°C + 273.15 = 298.15 K
6. Now we are ready to calculate the volume of the gas.
Using the moles of ethanol (0.040 mol), we can calculate the volume.
V = (0.040 mol) x (0.0821 L·atm/mol·K) x (298.15 K) / (1.00 atm)
Calculating this equation will give us the total volume of gas present after the reaction is complete.
The question asks for the TOTAL volume of gas which will be the CO2 produced PLUS the oxygen remaining that did not react.
CH3COOH + 2O2 ==> 2CO2 + 2H2O
mols CH3COOH = grams/molar mass = 1.86/60 = 0.031
That will use 0.031 x 2 = 0.062 mols O2.
mols O2 initially = 10.0 g/32 = 0.312
mols O2 unreacted = 0.312 - 0.062 = 0.25
mols CO2 formed = 0.031 x 2 = 0.062
The water @ 25 C will not be a gas.
Total mols gas = 0.25 mols O2 + 0.062 mols CO2 = 0.312 @ 273 K. You want the volume @ 298 K so use PV = nRT
Post your work if you get stuck.