Question

Acetylene (C2H2) is produced through the following reaction:

CaC2(s)+2H2O(1)>C2H2 (g)+Ca(OH)(aq)

A sample of 0.5 g C2H2 is collected over water at total pressure of 738 torr and 23C. The vapor pressure of water is 21 torr. What is the volume of C2H2 collected?

Answer 1

The total pressure is the sum of the partial pressures due to C₂H₂(g) and H₂O(g):

p(total) = 738 torr = p(H₂O) + p(C₂H₂) = 21 torr + p(C₂H₂)

p(C₂H₂) = 738 – 21 = 717 torr

Because 1 atm = 760 torr, p(C₂H₂) = 717/760 = 0.943 atm.

The molar mass of M(C₂H₂) is 26.036 g/mol.

n=m/M

n(C₂H₂) = 0.5 g / 26.036 (g/mol) = 0.0192 mol

Using the ideal gas law, pV = nRT, the volume of C₂H₂ collected

V = nRT/p

R = 0.08206 L×atm/mol×K

T = 23 + 273.15 = 296.15 K

V(C₂H₂) = (0.0192×0.08206×296.15)/0.943 = 0.494 L = 494 mL

Answer: 494 mL