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) →  $n = \frac{m}{MM} m (mass) = ? T = 24ºC Celsius to Kelvin TK = TºC + 273 TK = 24 + 273 TK = 297 By the equation of state of the gases or equation of Clapeyron, we have: [tex]P*V = n*R*T$
 
Since $n = \frac{m}{MM}$, we can perform the following substitution in the above Clapeyron equation:


$P * V = n * R * T$
$P*V = \frac{m}{MM} *R*T$
multiply cross
$m = \frac{P*V*MM}{R*T}$

Solving:
$m = \frac{P*V*MM}{R*T}$
$m = \frac{1.6*12.7*30}{0.082*297}$
$m = \frac{609.6}{24.354}$
$\boxed{\boxed{m \approx 25.03\:grams}}\end{array}}\qquad\quad\checkmark$

Answer:
25.03 grams of ethane gas