Question

(40 Points) Complete, balance, compute the amounts of the products assuming 100% yield. 12g Chlorine + 24g Ferric iodide.

Answer 1

The balanced reaction is :

`3Cl_(2)(g)+2FeI_(3)(aq)\rightarrow 2FeCl_(3)(aq)+3I_(2)`

Iodine produce = 20.92 grams

Ferric Chloride produced = 8.92 grams

Step-by-step explanation:

Molar mass : The amount of sustance present in 1 mole of it.

1 mole = Molar mass of the substance.

1 mole of Cl2 = 71 gram

3 mole of Cl2 = 71 x 3 = 213 gram

1 mole of I2 = 253.8 gram

3 mole of I2 = 3 x 253.8 = 761.4 gram

1 mole of FeCl3 = 162.2 gram

2 mole of FeCl3 = 2 x 162.2 = 324.4 gram

1 mole of FeI3 = 436.56 gram

2 mole of FeI3 = 2 x 436.56 gram =873.12

(use periodic table for calculation)

According to the given question 12 gram of Cl2 reacts with 24 gram of FeI3

The balanced equation is :

`3Cl_(2)(g)+2FeI_(3)(aq)\rightarrow 2FeCl_(3)(aq)+3I_(2)`

Here

`Cl_(2) =chlorine`

`FeI_(3)=ferric\ Iodide`

`FeCl3= ferric Chlorie`

`I_(2)= Iodine`

This shows:

3 mole Cl2 = 2 mole FeI3 = 2 mole FeCl3 = 3 mole I2

first ,calculate the limiting reagent (the substance present in less quantity)

3 mole Cl2 = 2 mole FeI3

213 gram of Cl2 = 873.12 gram of FeI3

1 gram =

`(873.12)/(213)`

12 gram of Cl2 will give =

`(873.12)/(213)* 12`

= 49.18 gram of FeI3 is needed

FeI3 present = 25 gram

So, FeI3 is limiting reagent as it is less than needed

Now , Find the amount of Products with 24 gram of FeI3 because it is the limiting reagent

1. To calculate FeCL3 produced

2 mole FeI3 = 2 mole FeCl3

873.12 gram of FeI3 = 324.4 gram of FeCl3

1 gram =

`(324.4)/(873.12)`

24 g FeI3 =

`(324.4)/(873.12)* 24` of FeCl3

= 8.92 gram of FeCl3

2. To calculate I2

2 mole of FeI3 = 3 mole I2

873.12 gram of FeI3 = 761.4 gram of I2

24 gram of FeI3

`(761.4)/(873.12)* 24`

= 20.92 gram of I2