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mote1985 [20]
3 years ago
10

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:2KClO3(s)Arrow.gif2KCl(s) + 3O2(g)T

he 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.
Chemistry
1 answer:
Eduardwww [97]3 years ago
5 0

Answer:

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:

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\times 9.49L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{730.5\times 9.49}{62.3637\times 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 \frac{2}{3}\times 0.3794 moles of potassium chlorate undergoes reaction.

<u>Moles of potassium chlorate reacted = 0.2529 moles</u>

(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\times 9.99L=n\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 293K\\\\n=\frac{735.5\times 9.99}{62.3637\times 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\times 32g/mol)=12.8675g

<u>Hence, the amount of oxygen gas collected will be 12.8675 g</u>

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