A1.52-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)

(a) 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

Explanation:

The reaction for the above condition is  

Fe(ClO4)3 (s) -> FeCl3 (s) + 6 O2 (g)

We have to calculate the number of moles of Fe(ClO4)3

First step  

  calculate the number of moles of Fe(ClO4)3

Given                        

Molar mass of Fe(ClO4)3 = 354.20g/mol

                 Given mass = 1.52 g

Number of moles of Fe(ClO4)3  = ?

Number of moles of Fe(ClO4)3 =   Given mass / Molar mass

                                      = 1.52/ 354.20  = 0.00429 moles of of Fe(ClO4)3

Second step  

   Calculation of number of moles of oxygen

As we know that from the reaction

1 mole of  Fe(ClO4)3   gives = 6 moles of oxygen

0.00429 moles of Fe(ClO4)3gives = 0.00429 x 6

                                                                  =0.0257 moles of oxygen  

Third step     Calculation of volume At STP

according to ideal gas law

given            Pressure =  1 atm                     Temperature = 273 k    

n number of moles= 0.0257 moles

R ideal gas constant= R = 0.0821 Latmmol-1K-1

             Volume =  ?

Ideal gas law=   PV=nRT

So solving for V= nRT/P

                         = 0.0257x0.0821x273/ 1

                         = 0.576 liter= 576 ml  

576 ml volume of oxygen gas should be produced under conditions of STP from a 1.52-g sample of Fe(ClO4)3.  

b.

the accuracy of the data.