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

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2