The balanced chemical reaction between Aluminum and Iodine is,
[tex]2Al(s) +3I_{2}(s)-->2AlI_{3}(s)[/tex]
Mass of Aluminum = 20.4 g
Moles of Aluminum = [tex]20.4 g Al * \frac{1molAl}{26.98g Al}=0.756molAl[/tex]
The stoichiometric mole ratio of iodine to Al: [tex]\frac{3molI_{2} }{2molAl}[/tex]
Moles of Iodine that would react with 0.756 mol Al=[tex]0.756molAl*\frac{3molI_{2} }{2molAl}=1.134molI_{2}[/tex]
Mass of iodine = [tex]1.134molI_{2}*\frac{253.81g I_{2} }{1 molI_{2} } =287.8g I_{2}[/tex]
Therefore, 287.8 g iodine would react with 20.4 g aluminum completely to form aluminum iodide.
287.8 g iodine would react with 20.4 g aluminum completely to form aluminum iodide.
Balanced chemical reaction:
[tex]2Al+3I_2---->2AlI_3[/tex]
Given:
Mass of Aluminum = 20.4 g
To find:
Moles of Aluminum = [tex]\frac{20.4g}{27g/mol}=0.756mol[/tex]
The stoichiometric mole ratio of iodine to Al: [tex]\frac{3mol I_2}{2 mol Al}[/tex]
Moles of Iodine that would react with 0.756 mol Al=[tex]0.756 mol*\frac{3 mol I_2}{2 mol Al}=1.134 mol I_2[/tex]
Mass of iodine = [tex]1.134mol I_2*\frac{253.81gI_2}{1mol I_2}=287.8gI_2[/tex]
Therefore, 287.8 g iodine would react with 20.4 g aluminum completely to form aluminum iodide.
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