Question

A 23.0-mL volume of O2, collected over water at 752 torr and 22°C, is produced from the thermal decomposition of KClO3. The vapor pressure of water at 22°C is 19.8 torr. How many moles of O2 are collected?

Answer 1

Answer:

$9.14\times 10^{-4}$ moles

Explanation:

We are given:

Vapor pressure of water = 19.8 torr

Total vapor pressure = 752 torr

Vapor pressure of oxygen gas = Total vapor pressure - Vapor pressure of water = (752 - 19.8) torr = 732.2 torr

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 = 732.2 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,  

Pressure = 732.2 / 760 atm = 0.9634 atm

V = Volume of the gas = 23 mL = 0.023 L

T = Temperature of the gas = $22^oC=[22+273]K=295K$

R = Gas constant = $0.0821 L.atm/K.mol$

n = number of moles of oxygen gas = ?

Applying the equation as:

0.9634 atm × 0.023 L = n × 0.0821 L.atm/K.mol × 295.15 K  

⇒n = $9.14\times 10^{-4}$ moles