Question

3. For the reaction whose balanced equation is as follows, find the number of

grams of Iz that will be formed when 300.0 9 of bromine react.
2 KI + Br2 → 2 KBr + I2

Answer 1

To find the number of grams of I2 formed, use the stoichiometry of the reaction and convert grams to moles using the molar mass of bromine (Br2). Then, convert moles to grams using the molar mass of iodine (I2).

Step-by-step explanation:

To determine the number of grams of I2 that will be formed when 300.0 g of bromine reacts, you can use the stoichiometry of the reaction. From the balanced equation, you can see that 1 mole of Br2 reacts to form 1 mole of I2. To convert grams to moles, use the molar mass of bromine (Br2).

Molar mass of Br2 = 2 x atomic mass of Br = 2 x 79.9 g/mol = 159.8 g/mol

Number of moles of Br2 = mass of Br2 / molar mass of Br2

= 300.0 g / 159.8 g/mol

= 1.88 moles of Br2

Since the reaction is 1:1 between Br2 and I2, the number of moles of I2 formed will also be 1.88 moles. To convert moles to grams, use the molar mass of I2.

Molar mass of I2 = 2 x atomic mass of I = 2 x 126.9 g/mol = 253.8 g/mol

Number of grams of I2 = moles of I2 x molar mass of I2

= 1.88 moles x 253.8 g/mol

= 476.9 g

Answer 2

The mass of iodine produced in grams is :

= 476.605 grams of I2

Step-by-step explanation:

Molar mass : It is the mass in gram present in 1 mole of the compound.

The atomic mass of Br = 79.904 g/mol

So , molar mass of Br2 = 2(mass of Br) = 2(79.04) = 159.808 gram

1 mole of Br2 = 159.808 gram

Atomic Mass of I = 126.904 gram

Mass of I2 = 2(mass of I) = 2(126.904)

Mass of I2 = 253.808 gram

1 mole of I2 = 253.808 gram

The balanced reaction is :

`2KI(s)+Br_(2)(l)\rightarrow 2KBr+I2`

According to this reaction ,

1 mole of Br2 will give = 1 mole of I2

159.808 gram of Br2 = 253.808 gram

So , 1 gram of Br2 =

`(253.808)/(159.808)` of I2

(Always calculate first for 1 gram by dividing )

300.09 gram of Bromine will produce =

`(253.808)/(159.808)* 300.09`

= 476.605 grams of I2