Question

How many g I, will react with 20.4 g Al to form AlI, (aluminum iodide)? As always, first balance the reaction! (Hint this is not a limiting reagent problem) A. 144. g B. 288. G C. 192. G D. 72 E. 576 g

Answer 1

B. 288. G

Step-by-step explanation:

Moles of Al :

Given, Mass of Al = 20.4 g

Molar mass of Al = 26.9815 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`Moles= (20.4\ g)/(26.9815\ g/mol)`

`Moles\= 0.7561\ mol`

The reaction between Al and I is shown below as:

2Al + 3I₂ ⇒ 2AlI₃

2 moles of aluminium react with 3 moles of iodine

1 mole of aluminium react with 3/2 moles of iodine

0.7561 moles of aluminium react with (3/2)*0.7561 moles of iodine

Moles of Iodine = 1.13415 moles

Molar mass of NaOH = 253.8089 g/mol

The formula for the calculation of moles is shown below:

`moles = (Mass\ taken)/(Molar\ mass)`

Thus,

`1.13415\ Moles= (Mass)/(253.8089\ g/mol)`

Mass of Iodine = 288 g