Question

A student collected the following data as part of the Determination of KClOx:

Temperature: 23.0 oC

Atmospheric Pressure: 749.3 torr

PH2O: 21.454 torr

Volume of O2: 89 mL

Mass of KClOx: 0.372 g

a. Determine the moles of O2 collected and the moles of monoatomic oxygen, O.

b. Determine the mass of O2 and of KCl at the end of the reaction.

c. Determine the mole ratio of O to KCl and provide the formula for the unknown

Answer 1

The equation for the decomposition of potassium chlorate is written as
2KClO3(s)→2KCl(s)+3O2(g)

To calculate the number of mol of O2 we use Gas Law:

PV = nRT
n = PV/RT
where,
P is the partial/vapor pressure O2
V is the volume of gas collected, 89 ml = 0.089 L
T is the temperature in Kelvin, 23.0 Oc = 296 K
R is the gas constant
n is the mass of O2 in mol.
given:
R = 0.08206 atm L / ºK mol
here

PH2O: 21.454 torr

tP = 749.3 torr -- the total pressure
solve for ppO2 - partial pressure O2
tP = ppH2O + ppO2
ppO2 = tP - ppH2O
ppO2 = 749.3 torr - 21.454 torr ppH2O
ppO2 = 727.846 torr ppO2

or 0.9576921051 atm

therefore,
n = PV/RT
n = (0.958)*0.089 / (0.08206*296)
n = 3.51*10^-3 mol O2

1 mole O2 = 2 mole O

3.51*10^-3 mol O2 = 7.02*10^-3 mol O

The equation for the decomposition of potassium chlorate is written as
2KClO3(s)→2KCl(s)+3O2(g)

Determine the mass of O2 and of KCl at the end of the reaction.

the mass of O2 = Molar mass * number of moles

= 31.99 g/ moles * 3.51*10^-3 mol O2

=0.112 g

Moles of KCl

3.51*10^-3 mol O2 * 2 mole KCl / 3 moles O2

= 2.34*10^-3 mole KCl

the mass of KCl = Molar mass * number of moles

= 2.34*10^-3 mole KCl* 74.5513 g/mol

=0.174 g

c. Determine the mole ratio of O to KCl and provide the formula for the unknown

the mole ratio of O to KCl = 7.02*10^-3 mol O / 2.34*10^-3 mole KCl

= 3

Thus the formulas is KClO3