Question

How many grams of ethane gas (C2H6) are in 12.7-liter sample at 1.6 atmosphere and 24 degrees Celsius? Show all work used to solve this problem.

Answer 1

Data:
M (Molar Mass) of C2H6
C = 2*12 = 24 amu
H = 6*1 = 6 amu
---------------------
MM C2H6 = 24+6 = 30 g/mol

P (pressure) = 1.6 atm
V (volume) = 12.7 L
R = 0,082 atm .L/mol.K
n (Number of mols) →
`image`

Since
`n = (m)/(MM)`, we can perform the following substitution in the above Clapeyron equation:



`P * V = n * R * T`

`P*V = (m)/(MM) *R*T`
multiply cross

`m = (P*V*MM)/(R*T)`

Solving:

`m = (P*V*MM)/(R*T)`

`m = (1.6*12.7*30)/(0.082*297)`

`m = (609.6)/(24.354)`

`\boxed{\boxed{m \approx 25.03\:grams}}\end{array}}\qquad\quad\checkmark`

Answer:
25.03 grams of ethane gas