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)

(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

Answer 1

Step-by-step 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.