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 748 mm Hg. If the wet C2H2 gas formed occupies a volume of 5.50 L, the number of moles of CaC2 reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

Answer : The number of moles of $CaC_2$ reacted was, 0.214 moles.

Explanation :

First we have to calculate the mole of $C_2H_2$ gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $C_2H_2$ gas = 748 mmHg - 23.8 mHg = 724.2 mmHg = 0.953 atm   (1 atm = 760 mmHg)

V = Volume of $C_2H_2$ gas = 5.50 L

n = number of moles $C_2H_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $C_2H_2$ gas = $25^oC=273+25=298K$

Putting values in above equation, we get:

$0.953atm\times 5.50L=n\times (0.0821L.atm/mol.K)\times 298K$

$n=0.214mol$

Now we have to calculate the moles of $CaC_2$

The balanced chemical reaction is:

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

From the balanced chemical reaction we conclude that,

As, 1 mole of $C_2H_2$ gas produced from 1 mole of $CaC_2$

So, 0.214 mole of $C_2H_2$ gas produced from 0.214 mole of $CaC_2$

Therefore, the number of moles of $CaC_2$ reacted was, 0.214 moles.