Question

A 1.52-gram sample of iron (III) perchlorate was decomposed according to the reaction below, producing 575 mL of oxygen gas at 298K and 761 torr. Fe(ClO4)3 (s) -> FeCl3 (s) + 6 O2 (g) what volume of oxygen gas should be produced under conditions of STP from a 1.52-g sample of Fe(ClO4)3? Show all work

Answer 1

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 $V_1=575 mL$

Pressure of the oxygen gas , $P_1= 761 Torr =\frac{761}{760 }atm=1.0013 atm$

(1 atm = 760 Torr)

Temperature of the gas of the oxygen gas ,$T_1=298 K$

At STP. the pressure of the gas = $P_2=1 atm$

At STP, the temperature of the gas= $T_2=273.15 K$

At STP. the volume of the oxygen gas = $V_2$

Using combined gas law to find out the volume of oxygen gas at STP:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}$

$V_2=\frac{1.0013 atm\times 575 mL\times 273.15 K}{298 K\times 1 atm}$

$V_2=527.74 mL$

527.74 mL of oxygen gas should be produced under conditions of STP from a 1.52 grams sample of iron (III) perchlorate .