Question

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)The product gas, O2, is collected over water at a temperature of 20 °C and a pressure of 748 mm Hg. If the wet O2 gas formed occupies a volume of 9.49 L, the number of moles of KClO3 reacted was ?mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)The product gas, O2, is collected over water at a temperature of 20 °C and a pressure of 753 mm Hg. If the wet O2 gas formed occupies a volume of 9.99 L, the number of grams of O2 formed is ?g. The vapor pressure of water is17.5 mm Hg at 20 °C.

Answer 1

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

Step-by-step explanation:

(a)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 748 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (748 - 17.5) mmHg = 730.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

`PV=nRT`

where,

P = pressure of the gas = 730.5 mmHg

V = Volume of the gas = 9.49 L

T = Temperature of the gas =
`20^oC=[20+273]K=293K`

R = Gas constant =
`62.3637\text{ L.mmHg }mol^(-1)K^(-1)`

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

`730.5mmHg* 9.49L=n* 62.3637\text{ L.mmHg }mol^(-1)K^(-1)* 293K\\\\n=(730.5* 9.49)/(62.3637* 293)=0.3794mol`

According to the reaction shown below as:-

`2KClO_3(s)\rightarrow 2KCl(s) +3O_2(g)`

3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.

So,

0.3794 mol of oxygen gas are produced when
`(2)/(3)* 0.3794` moles of potassium chlorate undergoes reaction.

Moles of potassium chlorate reacted = 0.2529 moles

(b)

We are given:

Vapor pressure of water = 17.5 mmHg

Total vapor pressure = 753 mmHg

Vapor pressure of Oxygen gas = Total vapor pressure - Vapor pressure of water = (753 - 17.5) mmHg = 735.5 mmHg

To calculate the amount of Oxygen gas collected, we use the equation given by ideal gas which follows:

`PV=nRT`

where,

P = pressure of the gas = 735.5 mmHg

V = Volume of the gas = 9.99 L

T = Temperature of the gas =
`20^oC=[20+273]K=293K`

R = Gas constant =
`62.3637\text{ L.mmHg }mol^(-1)K^(-1)`

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

`735.5mmHg* 9.99L=n* 62.3637\text{ L.mmHg }mol^(-1)K^(-1)* 293K\\\\n=(735.5* 9.99)/(62.3637* 293)=0.40211mol`

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

`0.037mol=\frac{\text{Mass of Oxygen gas}}{2g/mol}\\\\\text{Mass of Oxygen gas}=(0.40211mol* 32g/mol)=12.8675g`

Hence, the amount of oxygen gas collected will be 12.8675 g