Oxygen gas, generated by the reaction 2KClO3(s)->2KCl(s)+3O2(g) is collected over water at 27°C in a 1.55 L vessel at a total pressure of 1.00 ATM. (the vapor pressure of H2O at 27°C is 26.0 torr.) how many moles of KClO3 were consumed in the reaction?

Explanation:

Use Dalton's law and the vapor pressure of water at 23.0 o C to correct the pressure to units of atmoshperes.

PT = Poxygen +Pwater

At 23.0 o C the vapor pressure of water is 21.1 mmHg. (This can be found on a vapor pressure table.)

762 mmHg = Poxygen + 21.1 mmHg

Poxygen = 762 mmHg - 21.1 mmHg

Poxygen =741 mmHg

Convert the corrected pressure to atmospheres.

(741 mmHg) (1 atm / 760 mmHg) = 0.975 atm

Use the ideal gas law to find out how many moles of gas were produced:

PV = nRT (remember to put volume in liters and temperature in Kelvin)

(0.975 atm) (.193 L) = n (.0821 L atm / mol K) (298 K)

n = (0.975 atm) (.193 L) / (.0821 L atm / mol K) (298 K)

n = 7.69 X 10-4 mol

Use the number of moles and the molecular weight of oxygen to find out how many grams of oxygen were collected.

(7.69 X 10-4 mol) (32.0 g / 1 mol) = 2.46 X 10-2 g

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.