what is the volume occupied by 16.0g ethane gas (C2H6) at 720 Torr and 18C?

V=nRT/P

conversion of mol:
16.0g C2H6 X 1mol C2H6/30.07g C2H6= .53mol C2H6

V= .53 mol(0.0821 L*atm/mol*K)(291)K /
720 Torr (1 atm/760Torr)

V= 13.5 L

Here is a quick check. You have a half mole at near stp. About 11 liters. But in reality, your pressure is about ten percent lower than stp, so volume about 12 liters, and temp is about 5 percent higher, so volume is about 13 liters.

I would go with your answer.

To calculate the volume occupied by 16.0 g of ethane gas (C2H6) at 720 Torr and 18°C, we need to use the ideal gas law equation: V = nRT/P.

First, let's convert the mass of ethane (C2H6) to moles. The molar mass of C2H6 is 30.07 g/mol.

Mass of ethane (C2H6) = 16.0 g
Molar mass of ethane (C2H6) = 30.07 g/mol

Number of moles (n) = Mass / Molar mass
n = 16.0 g / 30.07 g/mol
n ≈ 0.532 mol (rounded to three decimal places)

Next, we need to convert the temperature from Celsius to Kelvin.
Temperature in Celsius (T) = 18°C

Temperature in Kelvin (K) = T + 273.15
K = 18°C + 273.15
K ≈ 291.15 K (rounded to two decimal places)

Now, we can substitute these values into the ideal gas law equation:
V = nRT / P

V = 0.532 mol * (0.0821 L * atm/mol*K) * 291.15 K / (720 Torr * (1 atm/760 Torr))

Simplifying the equation:
V ≈ 0.476 L

Therefore, the volume occupied by 16.0 g of ethane gas (C2H6) at 720 Torr and 18°C is approximately 0.476 L (rounded to three decimal places).

To find the volume occupied by 16.0 g of ethane gas (C2H6) at 720 Torr and 18°C, we can use the ideal gas law equation:

V = nRT / P

First, we need to convert the mass of ethane gas (C2H6) to moles using its molar mass. The molar mass of C2H6 is 30.07 g/mol.

16.0 g C2H6 * (1 mol C2H6 / 30.07 g C2H6) = 0.53 mol C2H6

Next, let's convert the temperature from Celsius to Kelvin.

18°C + 273.15 = 291 K

Now, we can substitute the values into the ideal gas law equation:

V = (0.53 mol) * (0.0821 L*atm/mol*K) * (291 K) / (720 Torr * (1 atm/760 Torr))

Simplifying the equation:

V = 0.53 mol * 0.0821 L*atm/mol*K * 291 K / (0.949 atm)

Finally, we can calculate the volume:

V = 0.53 mol * 0.0821 L * 291 K / 0.949

V ≈ 13.5 L

Therefore, the volume occupied by 16.0 g of ethane gas (C2H6) at 720 Torr and 18°C is approximately 13.5 L.