Problem 4 Consider the following reaction 4Fecl2aq 302g 2Fe

Problem 4 Consider the following reaction: 4Fecl2(aq) + 302(g) 2Fe2O3(s) + 4Cl2(g) What volume of a 0.680 M solution of FeCl2 is required to react completely with 6.43 x 1021 molecules of o2? mL Submit Answer Tries 0/3

Solution

Balanced equation:
4 FeCl₂ + 3 O₂ ===> 2 Fe₂O₃ + 4 Cl₂
Reaction type: single replacement

First let us calculate the moles of O2

Moles of O2 = 6.43 x 10²¹ / 6.023 x 10²³ = 0.01067 Moles

Moles of FeCl2 needed =  0.0142 Moles

Volume of FeCl2 = 0.0142 x 1000 / 0.68 = 20.93 ml

20.93 ml of FeCl2 is needed to react with 0.01067 moles of O2.