lron metal reacts with chlorne gas giving ironlil chloride T

lron metal reacts with chlorne gas giving iron(lil) chloride. The balanced chemical equation for this reaction is 2 Fe(s)-3 C12(g) 2 FeCl3(s) Beginning with 26.0 g of iron, what mass of Cl2, in grams, is required for complete reaction? What mass of FeCl3 can be produced? 49.5 gCl2 75.5 g FeCl Iron metal reacts with chlorine gas giving iron II) chloride. The balanced chemical equation for this reaction is: 2 Fe(s)-3C12(g) 2 FeCl3(s) 75.5 g FeCl3 can be produced from 26.0 g of iron and excess Cl2. If only 21.3 g of FeClj is obtained, what is the percent yield? d Iron metal reacts with chlorine gas giving iron(III) chloride. The balanced chemical equation for this reaction is : 2 Fe(s)-3C12(g) 2 FeCl3(s) If 26.0 g each of iron and chlorine are combined, what is the theoretical yield of iron(III) chloride?

Solution

2Fe(s) + 3Cl2(g) ------------> 2FeCl3(s)
2 moles of Fe react with 3 moles of Cl2
2*56g of Fe react with 3*71g of Cl2
26g of Fe react with = 3*71*26/2*56   = 49.5g
2 moles of Fe react with Cl2 to gives 2 moles of FeCl3
2*56g of Fe react with Cl2 to gives 2*162.2g of FeCl3
26g of Fe react with Cl2 to gives = 2*162.2*26/2*56 = 75.5g of FeCl3
c. actual yiled    = 21.3g

    theoretical yiled = 75.5g

   percnet yiled = actual yiled*100/theoretical yiled
                  = 21.3*100/75.5 = 28.2%

d. 2Fe(s) + 3Cl2(g) ------------> 2FeCl3(s)

no of moles Fe   = w/G.A.Wt

                             = 26/56 = 0.464moles

no of moles of Cl2   = W/G.M.Wt

                                = 26/71   = 0.366 moles

2 moles of Fe react with 3 moles of Cl2

0.464 moles of Fe react with = 3*0.464/2   = 0.696 moles of Cl2 is required.

Cl2 is limiting reactant

3 moles of Cl2 react with Fe to gives 2 moles of FeCl3

0.366 moles of Cl2 react with Fe to gives = 2*0.366/3   = 0.244 moles of FeCl3

mass of FeCl3 = no of moles * gram molar mass

                         = 0.244*162.2   = 39.57g of FeCl3

The theoretical yiled of FeCl3 = 39.57g