Answer to Question #166494 in General Chemistry for Teerrrrrrrrra

Question #166494

For the following reaction, 123 grams of iron are allowed to react with 268 grams of chlorine gas.


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


What is the FORMULA for the limiting reagent?...........


What is the maximum amount of iron(III) chloride that can be formed?  ...............grams


What amount of the excess reagent remains after the reaction is complete? ......... grams


please with steps



1
Expert's answer
2021-03-02T01:02:05-0500

Solution:

n(Fe) = 123/56 = 2.196 mole

n(FeCl3) = 2.196 mole

m(FeCl3) = 2.196×162 = 355.8 g

n(Cl2) = 268/71 = 3.775 mole

n(FeCl3) = 2.517 mole

m(FeCl3) = 2.517×162 = 407.7 g 


Thus Fe is limiting


The maximum amount of FeCl3 that can be formed = 407.7 g


To find amount of excess reagent (Cl2) remaining, we find how much was used up, and then subtract that from how much we started with.

n(Cl2) used up = 2.196 mole Cl2 × 3 mole Cl2 / 2 mole Fe = 3.294 mole Cl2 used

n(Cl2) remaining = 3.775 mole - 3.294 mole = 0.481 mole remaining

m(Cl2) remaining = 0.481 moles × 71 g/mol = 3.15 g Cl2 remain



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