Answer:
527.74 mL of oxygen gas should be produced under conditions of STP from a 1.52 grams sample of iron (III) perchlorate .
Explanation:
Volume of the oxygen gas under given conditions [tex]V_1=575 mL[/tex]
Pressure of the oxygen gas , [tex]P_1= 761 Torr =\frac{761}{760 }atm=1.0013 atm[/tex]
(1 atm = 760 Torr)
Temperature of the gas of the oxygen gas ,[tex]T_1=298 K[/tex]
At STP. the pressure of the gas = [tex]P_2=1 atm[/tex]
At STP, the temperature of the gas= [tex]T_2=273.15 K[/tex]
At STP. the volume of the oxygen gas = [tex]V_2[/tex]
Using combined gas law to find out the volume of oxygen gas at STP:
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
[tex]V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}[/tex]
[tex]V_2=\frac{1.0013 atm\times 575 mL\times 273.15 K}{298 K\times 1 atm}[/tex]
[tex]V_2=527.74 mL[/tex]
527.74 mL of oxygen gas should be produced under conditions of STP from a 1.52 grams sample of iron (III) perchlorate .
