Answer: 2588 grams of ethane are there in 100 L of this gas at 80 degree celsius and 25 atm.
Explanation:
According to the ideal gas equation:'
[tex]PV=nRT[/tex]
P = Pressure of the gas = 25 atm
V= Volume of the gas = 100 L
T= Temperature of the gas = 80°C = (80+273)K=353 K (0°C = 273 K)
R= Gas constant = 0.0821 atmL/K mol
n= moles of gas= ?
[tex]n=\frac{PV}{RT}=\frac{25\times 100}{0.0821\times 353}=86.3moles[/tex]
Mass of gas=[tex]moles\times {molar mass}}=86.3\times 30=2588g[/tex]
Thus 2588 grams of ethane are there in 100 L of this gas at 80 degree celsius and 25 atm.