Question

Calcium carbide reacts with water to produce acetylene gas according to the following equation: CaC2(s) + 2H2O(l)C2H2(g) + Ca(OH)2(aq) The product gas, C2H2, is collected over water at a temperature of 25 °C and a pressure of 749 mm Hg. If the wet C2H2 gas formed occupies a volume of 5.87 L, the number of moles of CaC2 reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C. Submit Answer

Answer 1

Answer: The number of moles of $CaC_2$ reacted was 0.229

Explanation:

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = Total pressure - vapor pressure of water = (749 - 23.8) mm Hg= 725.2 mm Hg = 0.954 atm    (760mmHg= 1atm)

V = Volume of gas = 5.87 L

n = number of moles = ?

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$25^0C=(25+273)K=298K$

$n=\frac{PV}{RT}$

$n=\frac{0.954atm\times 5.87L}{0.0821 L atm/K mol\times 298K}=0.229moles$

$CaC_2(s)+2H_2O(l)\rightarrow C_2H_2(g)+Ca(OH)_2(aq)$

According to stoichiometry:

1 mole of $C_2H_2$ is produced by = 1 mole of $CaC_2$

Thus 0.229 moles of $C_2H_2$ is produced by = $\frac{1}{1}\times 0.229=0.229$ moles of $CaC_2$

Thus number of moles of $CaC_2$ reacted was 0.229