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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)

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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

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