Respuesta :
Weight of CaC2 is 1.53 g
Temparature is 23 celcius T = 23 + 273K = 296K
The pressure of the gas P = 755 torr
Calculating the moles of CaC2 n = 1.53g x 64.0994 mol/g = 0.023869 moles
PV = nRT
V = nRT / P => V = (0.023869 mol) x (62.3636 L Torr / K mol) x (296 K) / 755 torr
Volume of C2H2 = 0.587 L
Answer:
0.60 L
Explanation:
Moles of [tex]CaC_2[/tex]:-
Mass = 1.53 g
Molar mass of [tex]CaC_2[/tex] = 64.099 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{1.53\ g}{64.099\ g/mol}[/tex]
[tex]Moles_{CaC_2}= 0.0239\ mol[/tex]
From the reaction shown below as:-
[tex]CaC_2_{(s)} + 2H_2O_{(g)}\rightarrow Ca(OH)_2_{(s)} + C_2H_2_{(g)}[/tex]
1 mole of [tex]CaC_2[/tex] on reaction forms 1 mole of [tex]C_2H_2[/tex]
0.0239 mole of [tex]CaC_2[/tex] on reaction forms 0.0239 mole of [tex]C_2H_2[/tex]
Mole of [tex]C_2H_2[/tex] = 0.0239 mol
Vapor pressure of water = 21.07 torr
Total vapor pressure = 755 torr
Vapor pressure of [tex]C_2H_2[/tex] = Total vapor pressure - Vapor pressure of water = (755 - 21.07) torr = 733.93 torr
To calculate the amount of hydrogen gas collected, we use the equation given by ideal gas which follows:
[tex]PV=nRT[/tex]
where,
P = pressure of the gas = 733.93 torr
V = Volume of the gas = ?
T = Temperature of the gas = [tex]25^oC=[23+273]K=296K[/tex]
R = Gas constant = [tex]62.3637\text{ L.torr }mol^{-1}K^{-1}[/tex]
n = number of moles of [tex]C_2H_2[/tex] = 0.0239 mol
Putting values in above equation, we get:
[tex]733.93torr\times V=0.0239 mol\times 62.3637\text{L.torr}mol^{-1}K^{-1}\times 296K\\\\V=\frac{0.0239\times 62.3637\times 296}{733.93}\ L=0.60\ L[/tex]
0.60 L is the volume of [tex]C_2H_2[/tex] that is collected over water.
