Question

Calculate the volume of c2h2 that is collected over water at 23 ∘c by reaction of 1.53 g of cac2 if the total pressure of the gas is 755 torr . (the vapor pressure of water is 21.07 torr.)

Answer 1

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 2

Answer:

0.60 L

Explanation:

Moles of $CaC_2$:-

Mass = 1.53 g

Molar mass of $CaC_2$ = 64.099 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{1.53\ g}{64.099\ g/mol}$

$Moles_{CaC_2}= 0.0239\ mol$

From the reaction shown below as:-

$CaC_2_{(s)} + 2H_2O_{(g)}\rightarrow Ca(OH)_2_{(s)} + C_2H_2_{(g)}$

1 mole of $CaC_2$ on reaction forms 1 mole of $C_2H_2$

0.0239 mole of $CaC_2$ on reaction forms 0.0239 mole of $C_2H_2$

Mole of $C_2H_2$ = 0.0239 mol

Vapor pressure of water = 21.07 torr

Total vapor pressure = 755 torr

Vapor pressure of $C_2H_2$ = 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:

$PV=nRT$

where,

P = pressure of the gas = 733.93 torr

V = Volume of the gas = ?

T = Temperature of the gas = $25^oC=[23+273]K=296K$

R = Gas constant = $62.3637\text{ L.torr }mol^{-1}K^{-1}$

n = number of moles of $C_2H_2$ = 0.0239 mol

Putting values in above equation, we get:

$733.93torr\times V=0.0239 mol\times 62.3637\text{L.torr}mol^{-1}K^{-1}\times 296K\\\\V=\frac{0.0239\times 62.3637\times 296}{733.93}\ L=0.60\ L$

0.60 L is the volume of $C_2H_2$ that is collected over water.