chemistry problem?

[匿名]

2015-09-25 8:25 pm

更新1:

For the following reaction, 23.9 grams of iron are allowed to react with with 36.4 grams of chlorine gas. iron (s) + chlorine (g) iron(III) chloride (s) What is the maximum amount of iron(III) chloride that can be formed? grams What is the FORMULA for the limiting reagent? What amount of the excess reagent remains after the reaction is complete?

回答 (1)

Roger the Mole

2015-09-25 11:30 pm

✔ 最佳答案

2 Fe + 3 Cl2 → 2 FeCl3

(23.9 g Fe) / (55.8452 g Fe/mol) = 0.42797 mol Fe
(36.4 g Cl2) / (70.9064 g Cl2/mol) = 0.51335 mol Cl2

0.51335 mole of Cl2 would react completely with 0.51335 x (2/3) = 0.34223 mole of Fe, but there is more Fe present than that, so Fe is in excess and Cl2 is the limiting reagent.

(0.51335 mol Cl2) x (2 mol FeCl3 / 3 mol Cl2) x (162.2040 g FeCl3/mol) = 55.5 g FeCl3

((0.42797 mol Fe initially) - (0.34223 mol Fe reacted)) x (55.8452 g Fe/mol) = 4.79 g Fe left over

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