Iodic acid, iron(II) iodide and hydrochloric acid react to form iron(III) chlori

Question

Iodic acid, iron(II) iodide and hydrochloric acid react to form iron(III) chloride, iodine monochloride and water. If 6.00 g of each reactant are mixed and allowed to react, what mass of iodine monochloride will form if the percent yield is 78.5% Write the equation using correct formulas for each substance. (1 part-all or nothing.) Balance the equation MUST SHOW WORK-all or nothing. What is the limiting reagent? (MUST SHOW WORK-all or nothing) What is the theoretical yield of ioding monochloride? What mass of iodine monochloride will actually form?

Explanation / Answer

Iodic acid :HIO3, Fe(Ii) iodide : FeI2 and Hydrochloric acid = HCl

HIO#+ FeI2+HCl ----->FeCl3+ ICL+ H2O

The balanced reactino is

5 HIO3 + 4 FeI2 + 25 HCl ---> 4 FeCl3 + 13 ICl + 15 H2O

theoretical molar rati of HIO3, FeI2 and HCl= 5: 4:25=   1.25 :1 : 6.25

molar masses : HIO3= 176, FeI2=310 HCl= 36.5

Moles : mass/Molecular weight

Moles in 6 gm ; HIO3= 6/176=0.034, FeI2= 6/310=0.019 and HCl= 6/36.5= 0.164

actual molar ratio = 0.034: 0.019 : 0.164=1.78: 1 : 4.82

limiting reactant is HCl. ( since stoichiomeric requirement is 6.25 and actual is 4.82)

from the reaction, 25 moles of HCl gvies 13 moles of ICl

0.164 moles gives 0.164*13/25= 0.0853 moles

Molecular weight of ICl= 162 , mass of ICl ( theoretical)= 0.0853*162= 13.82 gm

actual yiled= 75% of theoretical yield = 0.75*13.82 =10.4 gms

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