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

0.60 L

Step-by-step 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 = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (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* V=0.0239 mol* 62.3637\text{L.torr}mol^(-1)K^(-1)* 296K\\\\V=(0.0239* 62.3637* 296)/(733.93)\ L=0.60\ L`

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