A 45.0-L steel tank at 20.0°C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior?
A 45.0-L steel tank at 20.0°C contains acetylene gas, C2H2, at a pressure of 1.39 atm. Assuming ideal behavior, how many grams of acetylene are in the tank?
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Update:Please give answer because I don't even know what to do with the formula..
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Use the ideal gas equation
PV = nRT
we know that moles (n) = mass (m) / molar mass (M)
so substituting this into the ideal gas equation and we get.
PV = mRT / M
P = 1.39 atm
V = 45.0 L
m = ? g
M = 26.036 g/mol
R = gas constant = 0.028057 L atm mol^-1 K^-1
T = temperature in Kelvin = 20 deg C + 273.15 = 293.15 K
m = PVM / RT
= 1.39 atm x 45.0 L x 26.036 g/mol / (0.082057 x 293.15K)
= 67.7 g ( sig figs)
78.404 grams.