gram (g) of iron (III) chloride (FeCl3) can be made from 20 grams of chlorine and an 10 grams iron? Find the Excess and limiting reactant?

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  • 4 weeks ago

    2 Fe + 3 Cl2 → 2 FeCl3

    (20 g Cl2) / (70.9064 g Cl2/mol) = 0.28206 mol Cl2

    (10 g Fe) / (55.8450 g Fe/mol) = 0.17907 mol Fe

    0.17907 mole of Fe would react completely with 0.17907 x (3/2) = 0.268605 mole of Cl2, but there is more Cl2 present than that, so Cl2 is in excess and Fe is the limiting reactant.

    (0.17907 mol Fe) x (2 mol FeCl3 / 2 mol Fe) x (162.204 g FeCl3/mol) = 29 g FeCl3

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  • 4 weeks ago

    2 Fe + 3 Cl2 = 2 FeCl3 

    20 g of Cl2 = 20 g / 70.90 g/mol = 0.282 moles 

    these need 0.282 moles * 2/3 moles of Fe to react  = 0.188 mole

    we have 10g / 56 g/ mol of Fe =0.179 moles 

    the Fe is the limiting reagent ( 0.179 < 0.188) and forms 0.179 moles of FeCl3

    according to our equation 

    mass of these moles = 0.179 moles * 162.2040 g/mol = 29.0 g

    mass of FeCl3 formed = 29.0 g 

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