Question

Oxygen gas can be prepared by heating potassium chlorate according to the following equation:

2KClO3(s) → 2KCl(s) + 3O2(g)

The product gas, O2, is collected over water at a temperature of 20 °C and a pressure of 747 mm Hg. If the wet O2 gas formed occupies a volume of 6.42 L, the number of grams of O2 formed is _______g. The vapor pressure of water is 17.5 mm Hg at 20 °C.

Answer 1

Answer: The number of grams of
`O_2` formed is 8.32 g

Step-by-step explanation:

According to the ideal gas equation:

`PV=nRT`

P = Pressure of the gas = Total pressure - presure of water = (747-17.5)mm Hg= 729.5 mm Hg = 0.96 atm (760mmHg=1atm)

V= Volume of the gas = 6.42 L

n= moles of gas = ?

T= Temperature of the gas in kelvin =
`20^0C=(20+273)=293K`

R= Gas constant = 0.0821Latm/Kmol

`n=(PV)/(RT)=(0.96* 6.42)/(0.0821* 293)=0.26moles`

The balanced chemical reaction is:

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

Mass of
`O_2` produced=
`moles* {\text {Molar Mass}}=0.26mol* 32g/mol=8.32g`

Thus number of grams of
`O_2` formed is 8.32 g