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Question

For the following reaction,  18.9 grams of  iron are allowed to react with  40.3 grams of  chlorine gas .

iron(s) +  chlorine(g)   iron(III) chloride(s)

What is the maximum mass of  iron(III) chloride that can be formed?

What is the  FORMULA for the limiting reagent?

What mass of the excess reagent remains after the reaction is complete?

Answer 1

2Fe(s) + 3Cl₂(g) ===> 2FeCl₃(s) ... balanced equation

moles Fe = 18.9 g x 1 mol/55.85 g = 0.338 mol Fe (divided by 2 -> 0.169) = LIMITING REAGENT is Fe

moles Cl₂ = 40.3 g x 1 mol/71 g = 0.568 mol Cl₂ (divided by 3 -> 0.189)

Maximum mass FeCl₃ = 0.338 mol Fe x 2 mol FeCl3/2 mol Fe x 162 g/mol = 54.8 g FeCl₃ (3 sig. figs.)

Mass excess reagent remaining:

Started with 0.568 moles Cl₂

moles Cl₂ used = 0.338 mol Fe x 3 mol Cl₂/2 mol Fe = 0.507 moles Cl₂ used up

mass Cl₂ remaining = (0.568 mol - 0.507 mol) x 71 g/mol = 4.33 g Cl₂ remaining