You are given 2.677 g of a mixture of KClO3 and KCl. When heated, the KClO3 decomposes to KCl

and O2,
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g),
and 605 mL of O2 is collected over water at 28 °C. The total pressure of the gases in the collection flask is
763 mm Hg. What is the weight percentage of KClO3 in the sample?
The formula weight of KClO3 is 122.55 g/mol. The vapor pressure of water at 28 °C is 28.3 mm Hg

(763-28.3)(0.605)/(0.0821)....help?

PV = nRT

P = [(763-28.3)/760] atm.
V = 0.605 is correct.
Substitute R and T and solve for n = number of mols O2.

Convert mols O2 to mols KClO3 using the coefficients in the balanced equation. Then convert mols KClO3 to grams. g = mols x molar mass.
Then %KClO3 = (mass KClO3/2.677)*100 = ?

n= PV/RT

[(763-28.3)/760](0.605)/(8.314)(301.15)
=0.000233592 x (2/3) = 0.000155728 x 122.5 g
=0.019076722/2.677 x 100 = 0.710 %

This result seems really small...what have i done incorrect?

Yes, and it IS too small.

R = 0.08206 and not 8.314 when P is in atmospheres. Correct that and see if the answer is about 72%

To find the weight percentage of KClO3 in the sample, we need to first determine how much O2 gas was produced during the decomposition of KClO3.

From the reaction equation:
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)

We know that when 2 moles of KClO3 decompose, it produces 3 moles of O2. Therefore, the molar ratio between KClO3 and O2 is 2:3.

Given that 605 mL of O2 gas was collected over water at 28 °C, we need to convert this volume to moles using the ideal gas law equation:

PV = nRT

Where:
P = total pressure of the gases in the collection flask (763 mm Hg)
V = volume of O2 gas collected (605 mL = 0.605 L)
n = number of moles of O2 gas
R = ideal gas constant (0.0821 L·atm/(mol·K)) (Note: the units of R need to match the units of pressure and volume)
T = temperature of the gas in Kelvin (28 °C = 28 + 273 = 301 K)

Rearranging the ideal gas law equation:

n = PV / RT

Now we can calculate the number of moles of O2 gas produced.

n = (763 - 28.3) mm Hg * 0.605 L / (0.0821 L·atm/(mol·K) * 301 K)

Simplifying the equation gives:

n ≈ 19.73 mmol

Since the molar ratio between KClO3 and O2 is 2:3, we can calculate the number of moles of KClO3 by multiplying the number of moles of O2 by the ratio:

n(KClO3) = (2/3) * 19.73 mmol ≈ 13.15 mmol

Now we can calculate the weight of KClO3 in the sample:

weight(KClO3) = n(KClO3) * molar mass(KClO3)

Here the molar mass of KClO3 is given as 122.55 g/mol.

weight(KClO3) ≈ 13.15 mmol * 122.55 g/mol = 1607.33 mg

Finally, we can calculate the weight percentage of KClO3 in the sample:

weight percentage(KClO3) = (weight(KClO3) / weight of the mixture) * 100

Since we were given 2.677 g of the mixture, we can substitute the values into the formula:

weight percentage(KClO3) ≈ (1607.33 mg / 2677 mg) * 100 = 60.05%

Therefore, the weight percentage of KClO3 in the sample is approximately 60.05%.