# How many grams of ethane gas (C2H6) are in a 12.7 liter sample at 1.6 atmospheres and 24°C? Show all work used?

### 1 Answer

- JimLv 79 years agoFavorite Answer
28.05 Grams per Mole

PV = nRT

K = °C + 273.15, therefore 24C = 24 + 273.15 = 297.15K

1 atm = 101325 Pa = 101.325 kPa, therefore, 1.6 ATM = 1.6 x 101.325 kPa = 162.12 kPa

P=pressure

V=volume = 12.7L

n=moles of gas

T=temperature = 297.15K

R = gas constant (dependent on the units of pressure, temperature and volume)

R = 8.314 J K-1 mol-1 if

Pressure is in kilopascals(kPa) = 162.12 kPa

Volume is in litres(L)

Temperature is in kelvin(K)

R = 0.0821 L atm K-1 mol-1 if

Pressure is in atmospheres(atm)

Volume is in litres(L)

Temperature is in kelvinKelvin(K)

Solve the equation for n, number of moles of gas:

PV/RT = n

n = (162.12 kPa)(12.7L)/(8.314 J K-1 mol-1)(297.15K)

n = 2058.924/2470.5051

n = 0.8334 Moles

And since 1 mole of Ethane has a MW of 28.05 Grams, then

Grams of Ethane = 0.8334 x 28.05 = 23.38 Grams, approximately