Question

A trial of this decomposition experiment, using different quantities of reactants than those listed in the question above produced the following data: Volume of O2 produced at room conditions 280 mL Barometric pressure 740 torr Temperature of water 24°C Termperature of O2 25°C Vapor Pressure due to water at 25°C 22.4 torr For the conditions listed above, calculate the volume of O2(g) produced at standard conditions of temperature and pressure. (enter your answer in liters)

Answer 1

Answer : The volume of
`O_2(g)` produced at standard conditions of temperature and pressure is 0.2422 L

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

`(P_1V_1)/(T_1)=(P_2V_2)/(T_2)`

where,

`P_1` = initial pressure of
`O_2` gas = (740-22.4) torr = 717.6 torr

`P_2` = final pressure of
`O_2` gas at STP= 760 torr

`V_1` = initial volume of
`O_2` gas = 280 mL

`V_2` = final volume of
`O_2` gas at STP = ?

`T_1` = initial temperature of
`O_2` gas =
`25^oC=273+25=298K`

`T_2` = final temperature of
`O_2` gas =
`0^oC=273+0=273K`

Now put all the given values in the above equation, we get:

`(717.6torr* 280mL)/(298K)=(760torr* V_2)/(273K)`

`V_2=242.2mL=0.2422L`

Therefore, the volume of
`O_2(g)` produced at standard conditions of temperature and pressure is 0.2422 L