Answer to Question #166494 in General Chemistry for Teerrrrrrrrra

Question

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

Answer 1

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