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.

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Alleei Alleei · Answer 1 · 2020-04-02T11:36:14+02:00

Answer : The number of moles of [tex]CaC_2[/tex] reacted was, 0.214 moles.

Explanation :

First we have to calculate the mole of [tex]C_2H_2[/tex] gas.

Using ideal gas equation:

[tex]PV=nRT[/tex]

where,

P = Pressure of [tex]C_2H_2[/tex] gas = 748 mmHg - 23.8 mHg = 724.2 mmHg = 0.953 atm (1 atm = 760 mmHg)

V = Volume of [tex]C_2H_2[/tex] gas = 5.50 L

n = number of moles [tex]C_2H_2[/tex] = ?

R = Gas constant = [tex]0.0821L.atm/mol.K[/tex]

T = Temperature of [tex]C_2H_2[/tex] gas = [tex]25^oC=273+25=298K[/tex]

Putting values in above equation, we get:

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

[tex]n=0.214mol[/tex]

Now we have to calculate the moles of [tex]CaC_2[/tex]

The balanced chemical reaction is:

[tex]CaC_2(s)+2H_2O(l)\rightarrow C_2H_2(g)+Ca(OH)_2(aq)[/tex]

From the balanced chemical reaction we conclude that,

As, 1 mole of [tex]C_2H_2[/tex] gas produced from 1 mole of [tex]CaC_2[/tex]

So, 0.214 mole of [tex]C_2H_2[/tex] gas produced from 0.214 mole of [tex]CaC_2[/tex]

Therefore, the number of moles of [tex]CaC_2[/tex] reacted was, 0.214 moles.