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.

Moles of potassium chlorate reacted = 0.2529 moles

The amount of oxygen gas collected will be 12.8675 g

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:

where,

P = pressure of the gas = 730.5 mmHg  

V = Volume of the gas = 9.49 L

T = Temperature of the gas =

R = Gas constant =

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

According to the reaction shown below as:-

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 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:

where,

P = pressure of the gas = 735.5 mmHg  

V = Volume of the gas = 9.99 L

T = Temperature of the gas =

R = Gas constant =

n = number of moles of oxygen gas = ?

Putting values in above equation, we get:

Moles of Oxygen gas = 0.40211 moles

Molar mass of Oxygen gas = 32 g/mol

Putting values in above equation, we get:

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