During the combustion of ethane, C2H6(g), 40.4 grams of liquid are collected . How many liters of ethane, measured at 3.55 atm and 33 C, were burned?
Here is a link that will solve most stoichiometry problems.
http://www.jiskha.com/science/chemistry/stoichiometry.html
Use the above to calculate the grams ethane used the use PV = nRT to solve for volume ethane. The liquid to which they refer to is water from the following:
2C2H6 + 7O2 ==> 4CO2 + 6H2O
Post your work if you get stuck.
To find the volume of ethane burned, we need to use the ideal gas law equation: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
First, let's calculate the number of moles of ethane burned. We can use the molar mass of ethane (C2H6) to convert from grams to moles.
1. Calculate the molar mass of C2H6:
Molar mass of C: 12.01 g/mol
Molar mass of H: 1.01 g/mol
Molar mass of ethane (C2H6): (2 * 12.01 g/mol) + (6 * 1.01 g/mol) = 30.07 g/mol
2. Calculate the number of moles of ethane:
Moles = mass / molar mass
Moles = 40.4 g / 30.07 g/mol
Now, let's calculate the volume of ethane burned using the ideal gas law equation:
3. Convert the temperature from Celsius to Kelvin:
Kelvin = Celsius + 273.15
Kelvin = 33 °C + 273.15 = 306.15 K
4. Rearrange the ideal gas law equation to solve for volume:
V = (nRT) / P
5. Substitute the values into the equation:
V = (moles * R * T) / P
= (moles * 0.0821 L·atm/(mol·K) * 306.15 K) / 3.55 atm
6. Calculate the volume:
V = moles * 0.0821 L·atm/(mol·K) * 306.15 K / 3.55 atm
Now, plug in the calculated number of moles and solve for volume:
V = (40.4 g / 30.07 g/mol) * 0.0821 L·atm/(mol·K) * 306.15 K / 3.55 atm
After performing the calculations, you should get the volume of ethane burned, measured in liters.