Data:
M (Molar Mass) of C2H6
C = 212 = 24 amu
H = 61 = 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
