Exercise 10 | Properties of Gases

Acetylene C2H2 (g) is prepared by the reaction of calcium carbide with water:

CaC2 (s) + 2 H2O (l) → Ca(OH)2 (s) + C2H2 (g)


What is the volume of C2H2 (g) formed from 100.0 g of CaC2 (s) and 100.0 g of H2O at 25.0°C and 1.00 atm?

Limiting reagent:

nCaC2 = mCaC2 / MCaC2 = 100.0 / 64.1 = 1.56 mol

nH2O = mH2O / MH2O = 100.0 / 18.0 = 5.56 mol

2 moles of H2O react with 1 mole of CaC2 to form 1 mole of C2H2:

2 nCaC2 = 3.12 mol < 1 nH2O

⇒ CaC2 is the limiting reagent

 

From the chemical equation:

1 mole of CaC2 forms 1 mole of C2H2 ⇒ at the end of the reaction nC2H2 = nCaC2 (t =0) = 1.56 mol


Ideal gas equation:

PV = nRT

⇒ VC2H2 = nC2H2 RTPC2H2


V = volume in m3

n = molar mass in mol

R = ideal gas constant = 8.314 m3.Pa.K-1.mol-1

T = temperature in K ⇒ 25.0°C = 298 K

P = pressure in Pa ⇒ 1.00 atm = 1.013 x 105 Pa


VC2H2 = 1.56 × 8.314 × 2981.013 × 105 = 3.82 x 10-2 m3 = 38.2 L