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

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

$n=\frac{PV}{RT}=\frac{0.96\times 6.42}{0.0821\times 293}=0.26moles$

The balanced chemical reaction is:

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

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

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