Question

What mass of propane gas, C3H8, will occupy 593 mL at 904 mm Hg and 57 °C?

Answer 1

we have:

P = 904.0 mm Hg

= (904.0/760) atm

= 1.1895 atm

V = 593.0 mL

= (593.0/1000) L

= 0.593 L

T = 57.0 oC

= (57.0+273) K

= 330 K

find number of moles using:

P * V = n*R*T

1.1895 atm * 0.593 L = n * 0.08206 atm.L/mol.K * 330 K

n = 2.603*10^-2 mol

Molar mass of C3H8 = 3*MM(C) + 8*MM(H)

= 3*12.01 + 8*1.008

= 44.094 g/mol

we have below equation to be used:

mass of C3H8,

m = number of mol * molar mass

= 2.603*10^-2 mol * 44.094 g/mol

= 1.15 g

Answer: 1.15 g