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 25 °C and a pressure of 749 mm Hg. If the wet O2 gas formed occupies a volume of 5.76 L, the number of moles of KClO3 reacted was mol. The vapor pressure of water is 23.8 mm Hg at 25 °C.

Answer 1

For those looking for this problem's answer:

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

The product gas, O2, is collected over water at a temperature of 20 °C and a pressure of 748.0 mm Hg. If the wet O2 gas formed occupies a volume of 7.50 L, the number of moles of KClO3 reacted was ____ mol. The vapor pressure of water is 17.5 mm Hg at 20 °C.

Step-by-step explanation:

The answer is actually 2.00 mol. Same process as above just different numbers. Warning, y'all... DO NOT CHANGE FROM GRAMS TO MOLES!!! I made that mistake... Just take the number that you get from the equation and multiply it by the 2 mol KClO3/3 mol O2

Answer 2

The number of moles of KClO₃ reacted was 0,15 mol

Step-by-step explanation:

For the reaction:

2KClO₃(s) → 2KCl(s) + 2O₂(g)

The only gas product is O₂.

Total pressure is the sum of vapor pressure of water with O₂ gas formed. Thus, pressure of O₂ is:

749mmHg - 23,8mmHg = 725,2mmHg

Using gas law:

PV/RT = n

Where:

P is pressure (725,2mmHg ≡ 0,9542atm)

V is volume (5,76L)

R is gas constant (0,082 atmL/molK)

And T is temperature (25°C ≡ 298,15K)

Replacing, number of moles of O₂ are 0,2248 moles

As 2 moles of KClO₃ react with 3 moles of O₂ the moles of KClO₃ that reacted was:

0,2248 mol O₂×
`(2 mol KClO_(3))/(3 mol O_(2))` = 0,15 mol of KClO₃

I hope it helps!