If 237 mL of O2, measured at STP, are obtained by the decomposition of the KClO3 in a 5.44 g mixture of KCl and KClO3,

2KClO3(s)-->2KCl(s)+3O2(g)
what is the percent by mass of KClO3 in the mixture?

%KClO3 = (mass KClO3/mass sample) x 100 = ??

mass KClO3 = what?
0.237 g O2 x (1 mole O2/22.4L O2) x (2 moles KClO3/3 moles O2) x (122.6 g KClO3/mole KClO) = grams KClO3.

mass percent of oxygen in potassium chlorate:

To find the percent by mass of KClO3 in the mixture, we need to determine the mass of KClO3 and the total mass of the mixture.

Step 1: Calculate the molar mass of KCl, KClO3, and O2.
- The molar mass of KCl = (39.10 g/mol) + (35.45 g/mol) = 74.55 g/mol
- The molar mass of KClO3 = (39.10 g/mol) + (35.45 g/mol) + (3 * 16.00 g/mol) = 122.55 g/mol
- The molar mass of O2 = 2 * 16.00 g/mol = 32.00 g/mol

Step 2: Calculate the number of moles of O2.
- We can use the ideal gas law to calculate the number of moles of O2:
PV = nRT
Since the volume (V) is given in mL, we need to convert it to liters: V = 237 mL = 0.237 L
The pressure (P) is STP, which is 1 atm.
The gas constant (R) is 0.0821 L·atm/mol·K.
The temperature (T) is also STP, which is 273.15 K.
Solving for n (number of moles of O2), we get:
n = PV / RT = (1 atm * 0.237 L) / (0.0821 L·atm/mol·K * 273.15 K) ≈ 0.011 moles

Step 3: Calculate the number of moles of KClO3.
- According to the balanced equation, the molar ratio of KClO3 to O2 is 2:3.
- Since 2 moles of KClO3 produce 3 moles of O2, the number of moles of KClO3 is:
(0.011 moles O2) * (2 moles KClO3 / 3 moles O2) ≈ 0.0073 moles

Step 4: Calculate the mass of KClO3 and the total mass of the mixture.
- The mass of KClO3 = (0.0073 moles) * (122.55 g/mol) ≈ 0.894 g
- The total mass of the mixture is given as 5.44 g.

Step 5: Calculate the percent by mass of KClO3.
- The percent by mass of KClO3 = (mass of KClO3 / total mass of mixture) * 100%
= (0.894 g / 5.44 g) * 100% ≈ 16.44%

Therefore, the percent by mass of KClO3 in the mixture is approximately 16.44%.